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Unformatted text preview: COMMUTATIVE ALGEBRA PETE L. CLARK Contents Introduction 0.1. What is Commutative Algebra? 0.2. Why study Commutative Algebra? 0.3. Some themes of Commutative Algebra 1. Commutative rings 1.1. Fixing terminology 1.2. Adjoining elements 1.3. Ideals and quotient rings 1.4. Products of rings 1.5. A cheatsheet 2. Galois connections 2.1. The basics 2.2. Examples of Antitone Galois Connections 2.3. Examples of Isotone Galois Connections 3. Modules 3.1. Basic definitions 3.2. Finitely presented modules 3.3. Torsion and torsionfree modules 3.4. Tensor and Hom 3.5. Projective modules 3.6. Injective modules 3.7. Flat modules 3.8. Nakayama’s Lemma 3.9. Ordinal Filtrations and Applications 3.10. Tor and Ext 3.11. More on flat modules 3.12. Faithful flatness 4. First Properties of Ideals in a Commutative Ring 4.1. Introducing maximal and prime ideals 4.2. Radicals 4.3. Comaximal ideals 4.4. Local rings 5. Examples of rings 5.1. Rings of numbers 5.2. Rings of continuous functions 5.3. Rings of holomorphic functions 5.4. Polynomial rings 5.5. Semigroup algebras 1 4 4 4 7 7 7 10 11 14 16 17 17 18 20 20 20 25 28 28 30 37 45 46 50 57 65 71 74 74 77 79 82 83 83 84 90 92 94 2 PETE L. CLARK 6. Swan’s Theorem 6.1. Introduction to (topological) vector bundles 6.2. Swan’s Theorem 6.3. Proof of Swan’s Theorem 6.4. Applications of Swan’s Theorem 7. Localization 7.1. Definition and first properties 7.2. Pushing and pulling via a localization map 7.3. The fibers of a morphism 7.4. Commutativity of localization and passage to a quotient 7.5. Localization at a prime ideal 7.6. Localization of modules 7.7. Local properties 7.8. Local characterization of finitely generated projective modules 8. Noetherian rings 8.1. Chain conditions on posets 8.2. Chain conditions on modules 8.3. Semisimple modules and rings 8.4. Normal Series 8.5. The Krull-Schmidt Theorem 8.6. Some important terminology 8.7. Introducing Noetherian rings 8.8. The Bass-Papp Theorem 8.9. Artinian rings: structure theory 8.10. The Hilbert Basis Theorem 8.11. The Krull Intersection Theorem 8.12. Krull’s Principal Ideal Theorem 8.13. The Dimension Theorem 8.14. The Artin-Tate Lemma 9. Boolean rings 9.1. Definition and first properties of Boolean rings 9.2. Boolean Algebras 9.3. Ideal Theory in Boolean Rings 9.4. The Stone Representation Theorem 9.5. Boolean spaces 9.6. Stone Duality 9.7. Topology of Boolean Rings 10. Primary Decomposition 10.1. Some preliminaries on primary ideals 10.2. Primary decomposition, Lasker and Noether 10.3. Irredundant primary decompositions 10.4. Uniqueness properties of primary decomposition 10.5. Applications in dimension zero 10.6. Applications in dimension one 11. Affine k -algebras and the Nullstellensatz 11.1. Zariski’s Lemma 11.2. Hilbert’s Nullstellensatz 11.3. Other Nullstellens¨tze a 100 101 102 103 106 109 109 111 113 114 114 115 116 118 121 122 122 123 126 128 132 133 135 136 139 140 143 145 146 146 146 147 150 152 153 156 156 157 157 160 161 162 165 165 165 166 167 170 COMMUTATIVE ALGEBRA 12. Goldman domains and Hilbert-Jacobson rings 12.1. Goldman domains 12.2. Hilbert rings 12.3. Jacobson Rings 12.4. Hilbert-Jacobson Rings 13. The spectrum of a ring 13.1. The Zariski spectrum 13.2. Properties of the spectrum: quasi-compactness 13.3. Properties of the spectrum: separation and specialization 13.4. Irreducible spaces 13.5. Noetherian spaces 13.6. The map on spectra induced by a homomorphism 13.7. Hochster’s Theorem 13.8. The support of a module 13.9. Rank functions revisited 14. Integrality in Ring Extensions 14.1. First properties of integral extensions 14.2. Integral closure of domains 14.3. Spectral properties of integral extensions 14.4. Integrally closed domains 14.5. The Noether Normalization Theorem 14.6. A Little Invariant Theory 14.7. Abstract algebraic number theory I: Galois extensions of integrally closed domains 15. Factorization 15.1. Kaplansky’s Theorem (II) 15.2. Atomic domains, (ACCP) 15.3. EL-domains 15.4. GCD-domains 15.5. Polynomial rings over UFDs 15.6. Application: Eisenstein’s Criterion 15.7. Application: Determination of Spec R[t] for a PID R 15.8. Characterization of UFDs among Noetherian domains 15.9. Power series rings over UFDs 15.10. Nagata’s Criterion for UFDs 15.11. Auslander-Buchsbaum 16. Principal rings and B´zout domains e 16.1. Principal ideal domains 16.2. A word on Euclidean functions 16.3. Some structure theory of principal rings 16.4. B´zout domains e 17. Valuation rings 17.1. Basic theory 17.2. Ordered abelian groups 17.3. Connections with integral closure 17.4. Another proof of Zariski’s Lemma 17.5. Discrete valuation rings 18. Normalization theorems 3 170 171 174 175 175 176 176 177 178 180 182 184 184 185 186 188 188 190 192 194 195 197 201 202 203 204 205 206 208 212 213 214 214 215 218 219 219 221 223 225 226 226 228 231 232 233 236 4 PETE L. CLARK 18.1. The First Normalization Theorem 18.2. The Second Normalization Theorem 18.3. The Krull-Akizuki Theorem 19. Fractional ideals 19.1. Definition and first properties 19.2. Principal Fractional Ideals 19.3. Invertible Fractional Ideals and the Picard Group 19.4. More on invertible ideals 19.5. Divisorial ideals 19.6. The Divisor Class Group 20. Dedekind domains 20.1. Characterization in terms of invertibility of ideals 20.2. Ideal factorization in Dedekind domains 20.3. Local characterization of Dedekind domains 20.4. Factorization into primes implies Dedekind 20.5. Generation of ideals in Dedekind domains 20.6. Finitely generated modules over a Dedekind domain 21. Pr¨fer domains u 21.1. Characterizations of Pr¨fer Domains u 21.2. Modules over a Pr¨fer domain u 22. Structure of overrings 22.1. Introducing overrings 22.2. Overrings of Dedekind domains 22.3. Overrings of Pr¨fer Domains u 22.4. Kaplansky’s Theorem (III) 22.5. Every commutative group is a Picard group 23. Krull domains 24. Connections with model theory References 237 238 239 239 239 240 241 243 244 246 246 246 247 249 249 250 251 253 253 255 255 256 256 258 259 260 260 261 261 Introduction 0.1. What is Commutative Algebra? Commutative algebra is the study of commutative rings and attendant structures, especially ideals and modules. This is the only possible short answer I can think of, but it is not completely satisfying. We might as well say that Hamlet, Prince of Denmark is about a fictional royal family in late medieval Denmark and especially about the title (crown) prince, whose father (i.e., the King) has recently died and whose father’s brother has married his mother (i.e., the Queen). Informative, but not the whole story! 0.2. Why study Commutative Algebra? What are the purely mathematical reasons for studying any subject of pure mathematics? I can think of two: COMMUTATIVE ALGEBRA 5 I. Commutative algebra is a necessary and/or useful prerequisite for the study of other fields of mathematics in which we are interested. II. We find commutative algebra to be intrinsically interesting and we want to learn more. Perhaps we even wish to discover new results in this area. I think it is fair to say that most beginning students of commutative algebra can relate to the first reason: they need, or are told they need, to learn some commutative algebra for their study of other subjects. Indeed, commutative algebra has come to occupy a remarkably central role in modern pure mathematics, perhaps second only to category theory in its ubiquitousness, but in a different way. (Basic) category theory provides a common language and builds bridges between different areas of mathematics: it is something like a circulatory system. Commutative algebra provides core results that other results draw upon in a foundational way: it is something like a skeleton. The field of mathematics which most of all draws upon commutative algebra for its structural integrity is algebraic geometry, namely the study of geometric properties of manifolds and singular spaces which arise as the loci of solutions to systems of polynomial equations. In fact there is a hard lesson here: sometime in the 19th century algebraic geometry split off from complex function theory and dierential geometry as its own branch of mathematics and then burgeoned dramatically at the turn of the (20th!) century and the years immediately thereafter. But by 1920 or so the practitioners of the subject had found their way into territory in which “purely geometric” reasoning led to serious errors. In particular they had been making arguments about how algebraic varieties behave generically, but they lacked the technology to even give a precise meaning to the term. Thus the subject ultimately proved invertebrate and began to collapse under its own weight. Starting around 1930 there began a heroic shoring up process in which the foundations of the subject were recast with commutative algebraic methods at the core. In fact this was done several times over, in different ways, by Zariski, Weil, Serre and Grothendieck, among others. For the last 60 years or so it has been impossible to study algebraic geometry without knowing commutative algebra – a lot of commutative algebra. (For instance, more than is contained in these notes!) The other field of mathematics which draws upon commutative algebra in an absolutely essential way is algebraic number theory. One sees this from the beginning in that the Fundamental Theorem of Arithmetic is the assertion that the ring Z is a unique factorization domain (UFD), a fundamental commutative algebraic concept. Moreover number theory was one of the historical sources of the subject. Notably the concept of Dedekind domain came from Richard Dedekind’s numbertheoretic investigations. Knowledge of commutative algebra is not as absolutely indispensable for number theory (at least, not at the beginning) as it is for algebraic geometry, but such knowledge brings a great clarifying effect to the subject. In fact the interplay among number theory, algebraic geometry and commutative algebra flows in all directions. What Grothendieck did in the 1960s (with important 6 PETE L. CLARK contributions from Chevalley, Serre and others) was to create a single field of mathematics that encompassed commutative algebra and classical algebraic geometry – and, as a byproduct, algebraic number theory: namely the theory of schemes. As a result, most contemporary number theorists are also partly commutative algebraists and partly algebraic geometers: we call this more cosmopolitan take on the subject arithmetic geometry. But in fact there are plenty of other areas of mathematics that draw upon commutative algebra in slightly less dramatic but still important ways. To mention some which will show up in later in these notes: • Invariant theory. • Differential topology. • General topology. • Order theory. • Model theory. The task of providing a commutative algebraic foundation for algebraic geometry alone – or even the single, seminal text of R. Hartshorne – is a daunting one. Happily, this task has been completed by David Eisenbud (perhaps the leading contemporary mathematician working at the interface of commutative algebra and algebraic geometry) in his text [Eis]. This work is highly recommended. It is also 797 pages long, so contains enough material for 3 − 5 courses in the subject. It would be folly to try to improve upon, or even successfully imitate, Eisenbud’s work here, and I certainly have not tried. I myself am an arithmetic geometer (which, as I tried to explain above, is a sort of uppity kind of number theorist), so it is not surprising that these notes are skewed more towards number theory than most introductory texts on commutative algebra. However for the most part a respectful distance is maintained: we rarely discuss number theory per se but rather classes of rings that a number theorist would like: Dedekind domains, valuation rings, B´zout domains, and so forth.1 e But just much as I have included some material of interest to number theorists I have included material making connections to other branches of mathematics, especially connections which are less traditionally made in commutative algebra texts. In fact at several points I have digressed to discuss topics and theorems which make connections to other areas of mathematics: 1If I may be so bold as to actually play one of my cards, for the past several years I have been interested in developing a more commutative algebraic take on classical algebraic number theory, which would have the pleasant consequence of developing the function field case not just in parallel with the number field case but as a second specialization of a more general theory. The sucess of this task is nowhere guaranteed and at any rate is not very near at hand. This text is, in part, a very preliminary step in this direction. COMMUTATIVE ALGEBRA • • • • • • §X.X §X.X §X.X §X.X §X.X §X.X on on on on on on 7 Galois connections. rings of continuous functions. vector bundles and Swan’s Theorem. Boolean rings, Boolean spaces and Stone Duality. the topology of prime spectra, including Hochster’s Theorem. invariant theory, including the Shephard-Todd-Chevalley-Theorem. As a whole, general topological issues arise surprisingly often in these notes. This was not by design, but seems to point to a kinship between these two fields, aspects of which are cetainly known to the experts. On the other hand, I do find commutative algebra to be of interest unto itself, and I have tried to craft a sustained narrative in which various basic questions get explored in increasing depth and certain structures recur. Like any bad writer, I am not confident that I have made these themes apparent in the text itself, so in the following section I will explicitly point some of them out to you. 0.3. Some themes of Commutative Algebra. 1. Commutative rings 1.1. Fixing terminology. We are interested in studying properties of rings, generally commutative rings and invariably rings with unity. So let us begin by clarifying this terminology. By a general algebra R, we mean a triple (R, +, ·) where R is a set endowed with a binary operation + : R × R → R – called addition – and a binary operation · : R × R → R – called multiplication – satisfying the following: (CG) (R, +) is a commutative group, (D) For all a, b, c ∈ R, (a + b) · c = a · c + b · c, a · (b + c) = a · b + a · c. For at least fifty years, there has been agreement that in order for an algebra to be a ring, it must satisfy the additional axiom of associativity of multiplication: (AM) For all a, b, c ∈ R, a · (b · c) = (a · b) · c. A general algebra which satisfies (AM) will be called simply an algebra. A similar convention that is prevalent in the literature is the use of the term nonassociative algebra to mean what we have called a general algebra: i.e., a not necessarily associative algebra. A ring R is said to be with unity if there exists a multiplicative identity, i.e., an element e of R such that for all a ∈ R we have e · a = a · e = a. If e and e′ are two such elements, then e = e · e′ = e′ . In other words, if a unity exists, it is unique, and we will denote it by 1. 8 PETE L. CLARK A ring R is commutative if for all x, y ∈ R, x · y = y · x. With very few exceptions – which will be made clear! – we will be working always in the category of commutative rings with unity. In a sense which will be made precise shortly, this means that the identity 1 is regarded as a part of the structure of a ring, and must therefore be preserved by all homomorphisms of rings. Probably it would be more natural to study the class of possibly non-commutative rings with unity, since, as we will see, many of the fundamental constructions of rings give rise, in general, to non-commutative rings. But if the restriction to commutative rings (with unity!) is an artifice, it is a very useful one, since two of the most fundamental notions in the theory, that of ideal and module, become significantly different and more complicated in the non-commutative case. It is nevertheless true that many individual results have simple analogues in the noncommutative case. But it does not seem necessary to carry along the extra generality of non-commutative rings; rather, when one is interested in the non-commutative case, one can simply remark “Proposition X.Y holds for (left) R-modules over a noncommutative ring R.” Notation: Generally we shall abbreviate x · y to xy . Moreover, we usually do not use different symbols to denote the operations of addition and multiplication in different rings: it will be seen that this leads to simplicity rather than confusion. Group of units: Let R be a ring with unity. An element x ∈ R is said to be a unit if there exists an element y such that xy = yx = 1. Exercise X.X: a) Show that if x is a unit, the element y with xy = yx = 1 is unique, denoted x−1 . b) Show that if x is a unit, so is x−1 . c) Show that, for all x, y ∈ R, xy is a unit ⇐⇒ x and y are both units. d) Deduce that the units form a group, denoted R× , under multiplication. e) Give an example of an element x in a ring R for which there exists y such that xy = 1 but there does not exist an element z such that zx = 1. Example (Zero ring): Our rings come with two distinguished elements, the additive identity 0 and the multiplicative identity 1. Suppose that 0 = 1. Then for x ∈ R, x = 1 · x = 0 · x, whereas in any rin g 0 · x = (0 + 0) · x = 0 · x + 0 · x, so 0 · x = 0. In other words, if 0 = 1, then this is the only element in the ring. It is clear that for any one element set R = {0}, 0 + 0 = 0 · 0 = 0 endows R with the structure of a ring. We call this ring the zero ring. The zero ring exhibits some strange behavior, such that it must be explicitly excluded in many results. For instance, the zero element is a unit in the zero ring, which is obviously not the case in any nonzero ring. A nonzero ring in which every nonzero element is a unit is called a division ring. A commutative division ring is called a field. Let R and S be rings (with unity). A homomorphism f : R → S is a map COMMUTATIVE ALGEBRA 9 of sets which satisfies (HOM1) For all x, y ∈ R, f (x + y ) = f (x) + f (y ). (HOM2) For all x, y ∈ R, f (xy ) = f (x)f (y ). (HOM3) f (1) = 1. Note that (HOM1) implies f (0) = f (0 + 0) = f (0) + f (0), so f (0) = 0. Thus we do not need to explcitly include f (0) = 0 in the definition of a group homomorphism. For the multiplicative identity however, this argument only shows that if f (1) is a unit, then f (1) = 1. Therefore, if we did not require (HOM3), then for instance the map f : R → R, f (x) = 0 for all x, would be a homomorphism, and we do not want this. Exercise X.X: Suppose R and S are rings, and let f : R → S be a map satisfying (HOM1) and (HOM2). Show that f is a homomorphism of rings (i.e., satisfies also f (1) = 1) iff f (1) ∈ S × . A homomorphism f : R → S is an isomorphism if there exists a homomorphism g : S → R such that: for all x ∈ R, g (f (x)) = x; and for all y ∈ S , f (g (y )) = y . Exercise X.X: Let f : R → S be a homomorphism of rings. Show that TFAE: (i) f is a bijection. (ii) f is an isomorphism. Remark: In many algebra texts, an isomorphism of rings (or groups, etc.) is defined to be a bijective homomorphism, but this gives the wrong idea of what an isomorphism should be in other mathematical contexts (e.g. for topological spaces). Rather, having defined the notion of a morphism of any kind, one defines isomorphism in the way we have above. Exercise X.X.X: a) Suppose R and S are both rings on a set containing exactly one element. Show that there is a unique ring isomorphism from R to S . (This is a triviality, but explains why are we able to speak of the zero ring, rather than simply the zero ring associated to one element set. We will therefore denote the zero ring just by 0.) b) Show that any ring R admits a unique homomorphism to the zero ring. One says that the zero ring is the final object in the category of rings. Exercise X.X.X: Show that for any ring S there exists a unique homomorphism from the ring Z of integers to S . (One says that Z is the initial object in the category of rings.) A subring R of a ring S is a subset R of S such that (SR1) 1 ∈ R. (SR2) For all r, s ∈ R, r + s R, r − s ∈ R, and rs ∈ R. 10 PETE L. CLARK Here (SR2) expresses that the subset R is an algebra under the operations of addition and multiplication defined on S . Working, as we are, with rings with unity, we have to be a bit more careful: in the presence of (SR2) but not (SR1) it is possible that R either does not have a multiplicative identity or, more subtly, that it has a multiplicative identity which is not the element 1 ∈ S . An example of the first phenomenon is S = Z, R = 2Z. An example of the second is S = Z, R = 0. A more interesting example is S = Z × Z – i.e., the set of all ordered pairs (x, y ), x, y ∈ Z with (x1 , y1 ) + (x2 , y2 ) = (x1 + x2 , y1 + y2 ), (x1 , y1 ) · (x2 , y2 ) = (x1 x2 , y1 y2 ) – and R = {(0, y ) | y ∈ Z}. Then with the induced addition and multiplication from S , R is isomorphic to the ring Z and the element (0, 1) serves as a multiplicative identity on R which is different from the (always unique) multiplicative identity 1S = (1, 1), so according to our conventions R is not a subring of S . Notice that if R is a subring of S , the inclusion map R → S is an injective homomorphism of rings. Conversely, if ι : R → S is an injective ring homomorphism, then R ∼ ι(R) and ι(R) is a subring of S , so essentially we may use ι to view R = as a subring of S . The only proviso here is that this certainly depends on ι: in general there may be other injective homomorphisms ι : R → S which realize R as a different subset of S , hence a different subring. 1.2. Adjoining elements. Let ι : R → S be an injective ring homomorphism. As above, let us use ι to view R as a subring of S ; we also say that S is an extension ring of R and write S/R for this (note: this has nothing to do with cosets or quotients!) We wish now to consider rings T such that R ⊂ T ⊂ S ; such a ring T might be called a subextension of S/R or an intermediate ring. Let X = {xi } be a subset of S . Then the partially ordered set of all subrings of T containing R and X contains a bottom element, given (as usual!) by taking the intersection of all of its elements. (This partially ordered set is nonempty, since S is in it.) We call this the ring obtained by adjoining the elements of X to R. In the commutative case, we denote this ring by R[{xi }], for reasons that will become more clear when we discuss polynomial rings in §X.X . √ Example: Take R = Z, S = C. Then Z[i] = Z[ −1] is the smallest subring of √ C containing (Z and) −1. Example: Take R = Z, S = Q, let P be any set of prime numbers, and put 1 1 X = { p }p∈P . Then there is a subring ZP := Z[{ p }p∈P ] of Q. Exercise: Let P , Q be two sets of prime numbers. Show TFAE: (i) ZP ∼ ZQ . = (ii) ZP = ZQ . (iii) P = Q. Exercise: Show that every subring of Q is of the form ZP for some P . COMMUTATIVE ALGEBRA 11 Note well that the adjunction process R → R[X ] is defined only relative to some extension ring S of R, although the notation hides this. In fact, one of the recurrent themes of the subject is the expression of the adjunction process in a way which √ depends only on R itself. In the first example, this is √ achieved by identifying −1 with its minimal polynomial t2 + 1 and replacing Z[ −1] with the quotient ring Z[t]/(t2 + 1). The second example will eventually be turned around: we will be able to give an independent definition of ZP as a certain “ring of fractions” formed from Z and then Q will be the ring of fractions obtained by taking P to be the set of all prime numbers. Nevertheless, the existence of such turnabouts should not cause us to forget that adjunction is relative to √ extension; indeed forgetting this can lead to serious an trouble. For instance, if 3 2 is the unique real cube root of 2 and ζ3 is a primitive √ cube √ of unity, then the three complex numbers with cube 2 are z1 = 3 2, root √2 z2 = 3 2ζ3 and z3 = 3 2ζ3 . Each of the rings Q[z1 ], Q[z2 ], Q[z3 ] is isomorphic to the ring Q[t]/(t3 − 2), so all three are isomorphic to each other. But they are not the same ring: on the one hand Q[z1 ] is contained in R and the other two are not. √ More seriously Q[z1 , z2 , z3 ] = Q[ 3 2, ζ3 ], which strictly contains any one of Q[z1 ], Q[z2 ] and Q[z3 ]. 1.3. Ideals and quotient rings. Let f : R → S be a homomorphism of rings, and put I = f −1 (0) = {x ∈ R | f (x) = 0}. Then, since f is in particular a homomorphism of commutative groups (R, +) → (S, +), I is a subgroup of (R, +). Moreover, it enjoys both of the following properties: (LI) For all i ∈ I and y ∈ R, iy ∈ I . (RI) For all j ∈ I and x ∈ R, xj ∈ I . Indeed, f (xj ) = f (x)f (j ) = f (x) · 0 = 0 = 0 · f (y ) = f (i)f (y ) = f (iy ). In general, let R be a ring. An ideal is a subset I ⊂ R which is a subgroup of (R, +) (in particular, 0 ∈ I ) and which satisfies (LI) and (RI). Theorem 1. Let R be a ring, and let I be a subgroup of (R, +). TFAE: (i) I is an ideal of R. (ii) There exists a ring structure on the quotient group R/I making the additive homomorphism R → R/I into a homomorphism of rings. When these conditions hold, the ring structure on R/I in (ii) is unique, and R/I is called the quotient of R by the ideal I . Proof. Consider the group homomorphism q : R → R/I . If we wish R/I to be a ring in such a way so that q is a ring homomorphism, we need (x + I )(y + I ) = q (x)q (y ) = q (xy ) = (xy + I ). 12 PETE L. CLARK This shows that there is only one possible ring structure, and the only question is whether it is well-defined. For this we need that for all i, j ∈ I , (x + i)(y + j ) − xy = xj + iy + ij ∈ I . Evidently this holds for all x, y, i, j iff (LI) and (RI) both hold. Remark: If R is commutative, then of course there is no difference between (LI) and (RI). For a non-commutative ring R, an additive subgroup I satisfying condition (LI) but not necessarily (RI) (resp. (RI) but not necessarily (LI)) is called a left ideal (resp. a right ideal). Often one says two-sided ideal to emphasize that (LI) and (RI) both hold. Much of the additional complexity of the non-commutative theory comes from the need to distinguish between left, right and two-sided ideals. We do not wish to discuss such complexities here, so henceforth in this section we assume (except in exercises, when indicated) that our rings are commutative. Example: In R = Z, for any integer n, consider the subset (n) = nZ = {nx | x ∈ Z} of all multiples of n. This is easily seen to be an ideal.2 The quotient Z/nZ is the ring of integers modulo n. An ideal I R is called proper. Exercise X.X: Let R be a ring and I an ideal of R. Show that TFAE: (i) I ∩ R× ̸= ∅. (ii) I = R. Exercise X.X: a) Let R be a commutative ring. Show that R is a field iff R has exactly two ideals, 0 and R. b) Let R be a not necessarily commutative ring. Show that TFAE: (i) The only one-sided ideals of R are 0 and R. (ii) R is a division ring. c) For a field k and an integer n > 1, show that the matrix ring Mn (k ) has no two-sided ideals but is not a division ring. Exercise X.X: In contemporary undergraduate algebra texts, there is a trend to define the finite ring Z/nZ in a different and apparently simpler way: put Zn = {0, 1, . . . , n − 1}. For any integer x, there is a unique integer k such that x − kn ∈ Zn . Define a function mod n : Z → Zn by mod n(x) := x − kn. We then define + and · on Zn by x + y := mod n(x + y ), xy = mod n(xy ). Thus we have avoided any mention of ideals, equivalence classes, quotients, etc. Is this actually simpler? (Hint: how do we know that Zn satisfies the ring axioms?) For any commutative ring R and any element y ∈ R, the subset (y ) = yR = {xy | x ∈ R} is an ideal of R. Such ideals are called principal. A principal ideal ring is a commutative ring in which each ideal is principal. Exercise X.X: a) The intersection of an arbitrary family of (left, right or two-sided) ideals in a possibly non-commutative ring is a (left, right or two-sided) ideal. ∩ b) Let {Ii } be a set of ideals in the commutative ring R. Show that ∩ i has the iI following property: for any ideal J of R such that J ⊂ Ii for all i, J ⊂ i I . 2If this is not known and/or obvious to the reader, these notes will probably be too brisk. COMMUTATIVE ALGEBRA 13 Let R be a ring and S a subset of R. There is then a smallest ideal of R con∩ taining S , namely Ii , where Ii are all the ideals of R containing S . We call this the ideal generated by S . This is a “top-down” description; as usual, there is a complementary “bottom-up” description which is not quite as clean but often more useful. Namely, put ∑ ⟨S ⟩ := { ri si | ri ∈ R, si ∈ S } i.e., the set of all finite sums of an element of R times an element of S . Proposition 2. For a subset S of a commutative ring R, ⟨S ⟩ is an ideal, the intersection of all ideals containing S . Exercise X.X: Prove Proposition X.X. Exercise X.X: Let R be a ring. a) For ideals I and J of R, define I + J = {i + j | i ∈ I, j ∈ J }. Show that I + J = ⟨I ∪ J ⟩ is the smallest ideal containing both I and J . b) Extend part a) to any finite number of ideals I1 , . . . , In . c) Suppose {Ii } is an arbitrary set of ideals of I . Give an explicit description of the ideal ⟨Ii ⟩. Remark: The preceding considerations show that the collection of all ideals of a commutative ring R, partially ordered by inclusion, form a complete lattice. If I is an ideal in the ring R, then there is a correspondence between ideals J of R containing I and ideals of the quotient ring R/I , exactly as in the case of a normal subgroup of a group: Theorem 3. (Correspondence theorem) Let I be an ideal of a ring R, and denote the quotient map R → R/I by q . Let I (R) be the lattice of ideals of R, II (R) be the sublattice of ideals containing I and I (R/I ) the lattice of ideals of the quotient ring R/I . Define maps Φ : I (R) → I (R/I ), J → (I + J )/I, Ψ : I (R/I ) → I (R), J → q −1 (J ). Then Ψ ◦ Φ(J ) = I + J and Φ ◦ Ψ(J ) = J . In particular Ψ induces an isomorphism of lattices from I (R/I ) to II (R). Proof. For all the abstraction, the proof is almost trivial. For J ∈ I (R), we check that Ψ(Φ(J )) = Ψ(J + I (mod I )) = {x ∈ R | x + I ∈ J + I } = J + I ∈ II (R). Similarly, for J ∈ I (R/I ), we have Φ(Ψ(J )) = J . Remark: In fancier language, the pair (Φ, Ψ) give an isotone Galois connection between the partially ordered sets I (R) and I (R/I ). The associated closure operator Φ ◦ Ψ on I (R/I ) is the identity, whereas the closure operator Ψ ◦ Φ on I (R) carries each ideal J to the smallest ideal containing both J and I . The correspondence theorem will be a constant companion in our study of rings and ideals. As is common, we will often use the map Ψ to identify the sets I (R/I ) and II (R). 14 PETE L. CLARK Exercise X.X: Let I be an ideal of R and {Ji } be a set of ideals of R. Show that Φ preserves suprema and Ψ preserves infima: Φ(⟨Ji ⟩) = ⟨Φ(Ji )⟩ and ∩ ∩ Ψ( Ji ) = Ψ(Ji ). 1.4. Products of rings. Let R1 and R2 be rings. The Cartesian product R1 × R2 has the structure of a ring with “componentwise” addition and multiplication: (r1 , r2 ) + (s1 , s2 ) := (r1 + s1 , r2 + s2 ). (r1 , r2 ) · (s1 , s2 ) := (r1 s1 , r2 s2 ). Exercise X.X.X: a) Show that R1 × R2 is commutative iff both R1 and R2 are commutative. b) R1 × R2 has an identity iff both R1 and R2 do, in which case e := (e1 , e2 ) is the identity of R1 × R2 . As for any Cartesian product, R1 × R2 comes equipped with its projections π1 : R1 × R2 → R1 , | (r1 , r2 ) → r1 π2 : R1 × R2 → R2 , | (r1 , r2 ) → r2 . The Cartesian product X1 × X2 of sets X1 and X2 satisfies the following universal property: for any set Z and any maps f1 : Z → X1 , f2 : Z → X2 , there exists a unique map f : Z → X1 × X2 such that f1 = π1 ◦ f , f2 = π2 ◦ f . The Cartesian product R1 × R2 satisfies the analogous universal property in the category of rings: Exercise X.X: For rings R1 , R2 , S and ring homomorphisms fi : S → Ri , there exists a unique homomorphism of rings f : S → R1 × R2 such that fi = πi ◦ f . So the Cartesian product of R1 and R2 is also the product in the categorical sense. As with sets, we can equally well take the Cartesian product over an arbitrary indexed family of rings: if {Ri }i∈I is a family of rings, their Cartesian product ∏ i∈I Ri becomes a ring under coordinatewise addition and multiplication, and satisfies the universal property of the product. Details are left to the reader. It is natural to ask whether the category of rings has a direct sum as well. In other words, given rings R1 and R2 we are looking for a ring R together with ring homomorphisms ιi : Ri → R such that for any ring S and homomorphisms fi : Ri → S , there exists a unique homomorphism f : R → S such that fi = f ◦ ιi . We recall that in the category of abelian groups, the Cartesian product group G1 × G2 also the categorical direct sum, with ι1 : g → (g, 0) and ι2 : g → (0, g ). Since each ring has in particular the structure of an abelian group, it is natural to wonder whether the same might hold true for rings. However, the map ι1 : R1 → R1 × R2 does not preserve the multiplicative identity (unless R2 = 0), so is not a homomorphism of rings when identities are present. Moreover, even in the category of algebras, in order to satisfy the universal property on the underlying additive subgroups, the COMMUTATIVE ALGEBRA 15 homomorphism f is uniquely determined to be (r1 , r2 ) → f1 (r1 ) + f2 (r2 ), and it is easily checked that this generally does not preserve the product. We will see later that the category of rings does have direct sums in the categorical sense: the categorical direct sum of R1 and R2 is given by the tensor product R1 ⊗Z R2 . Now returning to the case of commutative rings, let us consider the ideal structure of the product R = R1 × R2 . It is easy to see that if I1 is an ideal of R1 , then I1 × {0} = {(i, 0) | i ∈ I } is an ideal of the product; moreover the quotient R/I1 is isomorphic to R1 /I1 × R2 . Similarly, if I2 is an ideal, {0} × I2 is an ideal of R2 . Finally, if I1 is an ideal of R1 and I2 is an ideal of R2 , then I1 × I2 := {(i1 , i2 ) |i1 ∈ I1 , i2 ∈ I2 } is an ideal of R. In fact we have already found all the ideals of the product ring: Proposition 4. Let R1 and R2 be commutative rings, and let I be an ideal of R := R1 × R2 . Put I1 := {r1 ∈ R1 | ∃r2 ∈ R2 | (r1 , r2 ) ∈ I }, I2 := {r2 ∈ R2 | ∃r1 ∈ R1 | (r1 , r2 ) ∈ I }. Then I = I1 × I2 = {(i1 , i2 ) | i1 ∈ I1 , i2 ∈ I2 }. Proof. Observe first that I1 × {0} and {0} × I2 are ideals of R contained in I . Indeed, if i1 ∈ I1 , then (i1 , r2 ) ∈ I for some r2 and then (i1 , 0) = (i1 , r2 ) · (1, 0), and similarly for I2 . Therefore I1 × I2 = (I1 × {0}) + ({0} × I2 ) ⊂ I. Conversely, if (x, y ) ∈ I , then (x, y ) = (x, 0)(1, 0) + (0, y )(0, 1) ∈ I1 × I2 . Remark: Another way to express the result is that, corresponding to a decomposition R = R1 × R2 , we get a decomposition I (R) = I (R1 ) × I (R2 ). Let us call a commutative ring R disconnected if there exists nonzero rings R1 , R2 such that R ∼ R1 × R2 , and connected otherwise.3 If R is disconnected, = then choosing such an isomorphism φ, we may put I1 = φ−1 (R1 × {0}) and I2 = φ−1 ({0} × R2 ). Evidently I1 and I2 are ideals of R such that I1 ∩ I2 = {0} and I1 × I2 = R. Conversely, if in a ring R we can find a pair of ideals I1 , I2 with these properties then it will follow from the Chinese Remainder Theorem (Theorem 126) that the natural map Φ : R → R/I2 × R/I1 , r → (r + I2 , r + I1 ) is an isomorphism. Now Φ restricted to I1 induces an isomorphism of groups onto R/I2 (and similarly with the roles of I1 and I2 reversed). We therefore have a distinguished element of I1 , e1 := Φ−1 (1). This element e1 is an identity for the multiplication on R restricted to I1 ; in particular e2 = e1 ; such an element is called an idempotent. In 1 3We will see later that there is a topological space Spec R associated to every ring, such that Spec R is disconnected in the usual topological sense iff R can be written as a nontrivial product of rings 16 PETE L. CLARK any ring the elements 0 and 1 are idempotents, called trivial; since e1 = Φ−1 (1, 0) – and not the preimage of (0, 0) or of (1, 1) – e1 is a nontrivial idempotent. Thus a nontrivial decomposition of a ring implies the presence of nontrivial idempotents. The converse is also true: Proposition 5. Suppose R is a ring and e is a nontrivial idempotent element of R: e2 = e but e ̸= 0, 1. Put I1 = Re and I2 = R(1 − e). Then I1 and I2 are ideals of R such that I1 ∩ I2 = 0 and R = I1 + I2 , and therefore R ∼ R/I1 × R/I2 is a = nontrivial decomposition of R. Exercise X.X: Prove Proposition X.X. Exercise X.X: Generalize the preceding discussion to decompositions into a finite number of factors: R = R1 × · · · × Rn . 1.5. A cheatsheet. Let R be a commutative ring. Here are some terms that we will analyze in loving detail later, but would like to be able to mention in passing whenever necessary. R is an integral domain if xy = 0 =⇒ x = 0 or y = 0. An ideal p of R is prime if the quotient ring R/p is an integral domain. Equivalently, p is an ideal such that xy ∈ p =⇒ x ∈ p or y ∈ p. An ideal m of R is maximal if it is proper – i.e., not R itself – and not strictly contained in any larger proper ideal. Equivalently, m is an ideal such that the quotient ring R/m is a field. R is Noetherian if it satisfies any of the following equivalent conditions:4 (i) For any nonempty set S of ideals of R, there exists I ∈ S which is not properly contained in any J ∈ S . (ii) There is no infinite sequence of ideals I1 I2 . . . in R. (iii) Every ideal of R is finitely generated. R is Artinian (or sometimes, an Artin ring) if the partially ordered set of ideals of R satisfies the descending chain condition: there is no infinite sequence of ideals I1 I2 . . .. If I and J are ideals of a ring R, we define the colon ideal5 (I : J ) = {x ∈ R | xJ ⊂ I }. (I : J ) is also an ideal. 4See Theorem 208 for a proof of their equivalence. 5The terminology is unpleasant and is generally avoided as much as possible. One should think of (I : J ) as being something like the “ideal quotient” I/J (which of course has no formal meaning). Its uses will gradually become clear. COMMUTATIVE ALGEBRA 17 2. Galois connections 2.1. The basics. Let (X, ≤) and (Y, ≤) be partially ordered sets. A map f : X → Y is isotone (or order-preserving) if for all x1 , x2 ∈ X , x1 ≤ x2 =⇒ f (x1 ) ≤ f (x2 ). A map f : X → Y is antitone (or order-reversing) if for all x1 , x2 ∈ X, x1 ≤ x2 =⇒ f (x1 ) ≥ f (x2 ). Let (X, ≤) and (Y, ≤) be partially ordered sets. An antitone Galois connection between X and Y is a pair of maps Φ : X → Y and Ψ : Y → X such that: (GC1) Φ and Ψ are both antitone maps, and (GC2) For all x ∈ X and all y ∈ Y , x ≤ Ψ(y ) ⇐⇒ Φ(x) ≥ y . Remark 1.1: There is a pleasant symmetry between X and Y in the definition. That is, if (Φ, Ψ) is a Galois connection between X and Y , then (Ψ, Φ) is a Galois connection between Y and X . Here is a fancy rephrasing: every Galois connection (Φ, Ψ) has a dual Galois connection (Ψ, Φ). Here is a good exercise for the reader trying to wrap her mind around Galois connections for the first time. I can (virtually) guarantee that somewhere in your past lies a self-dual Galois connection, i.e., with X = Y and Φ = Ψ. What is it? Many examples will be given later on. First we want to establish the fundamental properties of any Galois connection. Definition: If (X, ≤) is a partially ordered set, then a mapping f : X → X is called a closure operator if it satisfies all of the following properties: (C1) For all x ∈ X , x ≤ f (x). (C2) For all x1 , x2 ∈ X , x1 ≤ x2 =⇒ f (x1 ) ≤ f (x2 ). (C3) For all x ∈ X , f (f (x)) = f (x). Proposition 6. The mapping Ψ◦Φ is a closure operator on (X, ≤) and the mapping Φ ◦ Ψ is a closure operator on (Y, ≤). Proof. Symmetry considerations (c.f. Remark 1.1) show that it is enough to consider the mapping x → Ψ(Φ(x)) on X . If x1 ≤ x2 , then since both Φ and Ψ are antitone, we have Φ(x1 ) ≥ Φ(x2 ) and thus Ψ(Φ(x1 )) ≤ Ψ(Φ(x1 )): (C2).6 For x ∈ X , Φ(x) ≥ Φ(x), and by (GC2) this implies x ≤ Ψ(Φ(x)): (C1). Finally, for x ∈ X , applying (C1) to the element Ψ(Φ(x)) of X gives Ψ(Φ(x)) ≤ Ψ(Φ(Ψ(Φ(x)))). Conversely, we have Ψ(Φ(x)) ≤ Ψ(Φ(x)), so by (GC2) Φ(Ψ(Φ(x)) ≥ Φ(x), 6In other words, the composition of two antitone maps is isotone. 18 PETE L. CLARK and applying the order-reversing map Ψ gives Ψ(Φ(Ψ(Φ(x)))) ≤ Ψ(Φ(x)). Thus Ψ(Φ(x)) = Ψ(Φ(Ψ(Φ(x))). Corollary 7. The operators (Φ, Ψ) in a Galois Connection satisfy the following tridempotence properties: a) For all x ∈ X , ΦΨΦx = Φx. b) For all y ∈ X , ΨΦΨy = Ψy . Exercise: Prove Corollary 7. Proposition 8. Let (Φ, Ψ) be a Galois connection between partially ordered sets X and Y . Let X = Ψ(Φ(X )) and Y = Ψ(Φ(Y )). a) X and Y are precisely the subsets of closed elements of X and Y respectively. b) We have Φ(X ) ⊂ Y and Ψ(Y ) ⊂ X . c) Φ : X → Y and Ψ : Y → X are mutually inverse bijections. Proof. a) If x = Ψ(Φ(x)) then x ∈ X . Conversely, if x ∈ X , then x = Ψ(Φ(x′ )) for some x′ ∈ X , so Ψ(Φ(x))) = Ψ(Φ(Ψ(Φ(x′ )))) = Ψ(Φ(x′ )) = x, so X is closed. b) This is just a reformulation of Corollary 7. c) If x ∈ X and y ∈ Y , then Ψ(Φ(x)) = x and Ψ(Φ(y )) = y . We speak of the mutually inverse antitone bijections Φ : X → Y and Ψ : Y → X as the Galois correspondence induced by the Galois connection (Φ, Ψ). In applications, the content usually lies in determining the subsets X and Y . Having established the basic results, we will now generally abbreviate the closure operators Ψ ◦ Φ and Φ ◦ Ψ to x → x and y → y . 2.2. Examples of Antitone Galois Connections. If X is a set, then the power set 2X (of all subsets of X ) is partially ordered via containment: for Y1 , Y2 ⊂ X , put Y1 ≤ Y2 if Y1 ⊂ Y2 . If A ⊂ 2X is any subset, then it gets an induced ordering by containment. In all of our examples of Galois connections below, the partially ordered sets X and Y are both of this form. Example 1 (Galois connection of a G-set): Let G be a group acting on a set X . Let X = 2X be the collection of all subsets of X and G be the collection of all subgroups of G (each made into a partially ordered set by inclusion, as above). We define Φ : X → G by Φ : A ⊂ X → {g ∈ G | ∀x ∈ A, gx = x} and Ψ : G → X by Ψ : H ⊂ G → X H = {x ∈ X | ∀g ∈ H, gx = x}. Let us check that (Φ, Ψ) is a Galois connection. It is easy to see (and probably familiar) that Φ and Ψ are antitone: a smaller set has a larger subgroup acting trivially on it, and a smaller subgroup has a larger set of fixed points. Now let Y COMMUTATIVE ALGEBRA 19 be a subset of X and H a subgroup of G, and let us unpack the condition that A ≤ Ψ(H ) ⇐⇒ Φ(A) ≥ H . In this case, A ≤ Ψ(H ) says that for every y ∈ A and every h ∈ H , hy = y , whereas Φ(A) ≥ H says that for every h ∈ H and every y ∈ A, hy = y . These two statements are plainly equivalent! Subexample 1.1: Let L/K be a field extension, X = L and G = Aut(L/K ). Then it is not hard to show that every finite subgroup of G is closed. On the other hand, every closed subset of X is a subextension F of L/K , but it is not necessarily the case that every subextension is closed. If L/K is finite, then so is G and thus every subgroup of G is closed. On the other side, much of the content of classical Galois theory resides in the fact that every subextension of L/K is closed iff L/K is normal and separable. In the case of L/K infinite algebraic, it is again the case that every subextension F of L/K is closed iff L/K is normal and separable. However, it is no longer the case that every subgroup of Aut(L/K ) is closed. Rather, the closed subgroups of Aut(L/K ) form a local base at the identity for a profinite topology on Aut(L/K ), the Krull topology. Example 2 (Galois connection on a set of functions) Let X and R be sets and S a subset of R. Let F ⊂ Map(X, R) be a set of functions f : X → R. We define a Galois connection (Φ, Ψ) between 2X and 2F . For V ⊂ X , we define Φ(V ) = {f ∈ F | f (V ) ⊂ S }. For I ⊂ F , we define Ψ(I ) = {x ∈ X |∀f ∈ I, f (x) ∈ S }. Again it is easy to see that Φ and Ψ are antitone. For V ⊂ X and I ⊂ F , the condition V ⊂ Ψ(I ) means that for all x ∈ V and all f ∈ I , f (x) ∈ S , whereas the condition I ⊂ Φ(V ) means that for all f ∈ I and all x ∈ V , f (x) ∈ S . These two statements are plainly equivalent! Suppose that R is an abelian group and S is a subgroup of R. Then Map(X, R) is an abelian group under pointwise addition. Further suppose that F is a subgroup of Map(X, R). Then for each V ⊂ X , Φ(V ) is the subgroup of all functions which map V into S . In particular, if S = {0}, then Φ(V ) is the subgroup of all functions vanishing at every point of V . Similarly, suppose that R is a commutative ring and S is an ideal of R. Then Map(X, R) is a commutative ring under pointwise addition and multiplication. Further suppose that Y is a subring of Map(X, R). Then for each V ⊂ X , Φ(V ) is the set of all functions f in F which map V into S : this is an ideal of Y . Subexample 2.1: K a field, X = K n , R = K , S = 0, F = K [x1 , . . . , xn ]. This is the Galois correspondence between ideals in K [x1 , . . . , xn ] and affine subvarieties of K n . It will be discussed in more detail in §X.X . Subexample 2.2: X a compact topological space, R = R, S = 0, Y the set of all continuous functions from X to R. This will be discussed in more detail in §X.X . 20 PETE L. CLARK Example 3: Let V be a K -vector space and ⟨, ⟩ a bilinear form on K . Let X = Y = 2V , and define a Galois connection (Φ, Ψ) by S ⊂ X → S ⊥ = {x ∈ X | ∀s ∈ S ⟨s, x⟩ = 0} and T ⊂ Y → ⊥ T = {x ∈ X | ∀t ∈ T, ⟨x, t⟩ = 0}. Note that if the bilinear form is symmetric than this is an example of a Galois connection with Φ = Ψ. Subexample 3.1: Let ⟨, ⟩ be a symmetric bilinear form on a finite dimensional vector space V over a field K of characteristic different from 2. Then every subspace of V is closed iff {0} is closed iff the form is nondegenerate. Example 4: Let L be a language, let X be the set of L-theories, and let Y be the class of all classes C of L-structures, each partially ordered by inclusion.7 For a theory T , we define Φ(T ) = CT to be the class of all models of T , whereas for a class C , we define Ψ(C ) to be the collection of all sentences φ such that for all X ∈ C , X |= φ. Exercise X.X (Covering Spaces): Let π : X ′ → X be a covering map of pathconnected topological spaces. Let G = Deck(X ′ /X ) the be the group of homeomorphisms σ : X ′ → X ′ such that π ◦ σ = π . a) Define a Galois connection between the subgroups of G and the subcovering spaces πY : Y → X of π . b) What is the induced Galois correspondence? c) How does the situation simplify when X ′ is the universal cover of X ? 2.3. Examples of Isotone Galois Connections. COMPLETE ME! ♣ 3. Modules 3.1. Basic definitions. Suppose (M, +) is an abelian group. For any m ∈ M and any integer n, one can make sense of n • m. If n is a positive integer, this means m + · · · + m (n times); if n = 0 it means 0, and if n is negative, then n • m = −(−n) • m. Thus we have defined a function • : Z × M → M which enjoys the following properties: for all n, n1 , n2 ∈ Z, m, m1 , m2 ∈ M , we have (ZMOD1) (ZMOD2) (ZMOD3) (ZMOD4) 1 • m = m. n • (m1 + m2 ) = n • m1 + n • m2 . (n1 + n2 ) • m = n1 • m + n2 • m. (n1 n2 ) • m = n1 • (n2 • m) It should be clear that this is some kind of ring-theoretic analogue of a group action on a set. In fact, consider the slightly more general construction of a monoid (M, ·) acting on a set S : that is, for all n1 , n2 ∈ M and s ∈ S , we require 1 • s = s and (n1 n2 ) • s = n1 • (n2 • s). 7Here we are cheating a bit by taking instead of a partially ordered set, a partially ordered class. We leave it to the interested reader to devise a remedy. COMMUTATIVE ALGEBRA 21 For a group action G on S , each function g • : S → S is a bijection. For monoidal actions, this need not hold for all elements: e.g. taking the natural multiplication action of M = (Z, ·) on S = Z, we find that 0• : Z → {0} is neither injective nor surjective, ±1• : Z → Z is bijective, and for |n| > 1, n• : Z → Z is injective but not surjective. Exercise: Let • : M × S → S be a monoidal action on a set. Show that for each unit m ∈ M – i.e., an element for which there exists m′ with mm′ = m′ m = 1 – m• : S → S is a bijection. Then the above “action” of Z on an abelian group M is in particular a monoidal action of (Z, ·) on the set M . But it is more: M has an additive structure, and (ZMOD2) asserts that for each n ∈ Z, n• respects this structure – i.e., is a homomorphism of groups; also (ZMOD3) is a compatibility between the additive structure on Z and the additive structure on M . These axioms can be restated in a much more compact form. For an abelian group M , an endomorphism of M is just a group homomorphism from M to itself: f : M → M . We write End(M ) for the set of all endomorphisms of M . But End(M ) has lots of additional structure: for f, g ∈ End(M ) we define f + g ∈ End(M ) by (f + g )(m) := f (m) + g (m), i.e., pointwise addition. We can also define f · g ∈ End(M ) by composition: (f · g )(m) := f (g (m)). Proposition 9. For any abelian group M , the set End(M ) of group endomorphisms of M , endowed with pointwise addition and multiplication by composition, has the structure of a ring. Exercise: Prove Proposition X.X. Exercise: Show that End(Z) = Z, and for any n ∈ Z, End(Z/nZ) = Z/nZ. (More precisely, find canonical isomorphisms.) These simple examples are potentially misleading: we did not say that the multiplication was commutative, and of course there is no reason to expect composition of functions to be commutative. Exercise X.X: a) Show that End(Z/2Z ⊕ Z/2Z) = M2 (Z/2Z), the (noncommutative!) ring of 2 × 2 matrices with Z/2Z-coefficients. b) If M is an abelian group and n ∈ Z+ , show End(M n ) = Mn (End(M )). Now observe that the statement that the action of Z on M satisfes (ZMOD1) through (ZMOD4) is equivalent to the following much more succinct statement: For any abelian group M , the map n ∈ Z → (n•) : M → M is a homomorphism of rings Z → End(M ). This generalizes very cleanly: if R is any ring (not necessarily commuative) and M is an abelian group, a homomorphism • : R → End(M ) will satisfy: for all 22 PETE L. CLARK r ∈ R, m, m1 , m2 ∈ M : (LRMOD1) (LRMOD2) (LRMOD3) (LRMOD4) 1 • m = m. r • (m1 + m2 ) = r • m1 + r • m2 . (r1 + r2 ) • m = r1 • m + r2 • m. (r1 r2 ) • m = r1 • (r2 • m). The terminology here is that such a homomorphism r → (r•) is a left R-module structure on the abelian group M . What then is a right R-module structure on M ? The pithy version is that it is a ring homomorphism from Rop , the opposite ring of R to End(M ). This definition makes clear (only?) that if R is commutative, there is no difference between left and right R-module structures. Since our interest is in the commutative case, we may therefore not worry too much. But for the record: Exercise X.X.X: Show that a homomorphism Rop → End(M ) is equivalent to a mapping • : M × R → M satisfying m • 1 = m, (m1 + m2 ) • r = m1 • r + m2 • r, m • (r1 + r2 ) = m • r1 + m • r2 , m • (r1 r2 ) = (m • r1 ) • r2 . As usual for multiplicative notation, we will generally suppress the bullet, writing rm for left R-modules and mr for right R-modules. The calculus of left and right actions is at the same time confusing and somewhat miraculous: it is a somewhat disturbing example of a purely lexicographical convention that has – or looks like it has – actual mathematical content. Especially, suppose we have an abelian group M and two rings R and S , such that M simultaneously has the structure of a left R-module and a right S -module. Thus we wish to entertain expressions such as rms for m ∈ M , r ∈ R, s ∈ S . But as stands this expression is ambiguous: it could mean either (r • m) • s or r • (m • s). We say that M is an R-S bimodule if both of these expressions agree. Here is what is strange about this: lexicographically, it is an associativity condition. But “really” it is a commutativity condition: it expresses the fact that for all r ∈ R, s ∈ S , (r•) ◦ (•s) = (•s) ◦ (r•): every endomorphism coming from an element of R commutes with every endomorphism coming from an element of S . Thus for instance: Exercise: Show that any ring R is naturally a left R-module and a right R-module. We will not deal with bimodules further in these notes. In fact, when we say COMMUTATIVE ALGEBRA 23 R-module at all, it will be understood to mean a left R-module, and again, since we shall only be talking about commutative rings soon enough, the distinction between left and right need not be made at all. Definition: For M a left R-module, we define its annihilator ann(M ) = {r ∈ R | ∀m ∈ M, rm = 0}. Equivalently, ann(M ) is the set of all r such that r· = 0 ∈ End(M ), so that it is precisely the kernel of the associated ring homomorphism R → End(M ). It follows that ann(M ) is an ideal of R (note: two-sided, in the noncommutative case). Definition: A left R-module M is faithful if ann(M ) = 0. Explicitly, this means that for all 0 ̸= r ∈ R, there exists m ∈ M such that rm ̸= 0. Exercise X.X: Let M be an R-module. Show that M has the natural structure of a faithful R/ ann(M )-module. Definition: Let M be a left R-module. A submodule of M is a subgroup N of (M, +) such that RN ⊂ N . The following result is extremely easy and allimportant: Theorem 10. Let R be a ring. The left R-submodules of R are precisely the left ideals of R. Exercise X.X: Prove Theorem 10. Definition: Let M and N be left R-modules. A homomorphism of R-modules is a homomorphism of abelian groups f : M → N such that for all r ∈ R, m ∈ M, n ∈ N , f (rm) = rf (m). Exercise X.X: a) Define an isomorphism of R-modules in the correct way, i.e., not as a bijective homomorphism of R-modules. b) Show that a homomorphism of R-modules is an isomorphism iff it is bijective. If N is a submodule of a left R-module M , then the quotient group M/N has a natural R-module structure. More precisely, there is a unique left R-module structure on M/N such that the quotient map M → M/N is a homomorphism of R-modules. (Exercise!) Exercise: Let I be a two-sided ideal of the ring R, so that the quotient ring R/I has the structure of a left R-module. Show that ann(R/I ) = I. In particular, every two-sided ideal of R occurs as the annihilator of a left R-module. Exercise X.X: a) Let⊕ be a ring and {Mi }i∈I a family of R-modules. Consider the R abelian group M = i∈I Mi . Show that putting r(mi ) = (rmi ) makes R into an R-module. Show that the usual inclusion map ιi : Mi → M is a homomorphism of R-modules. b) Show that for any R-module N and R-module maps fi : Mi → N , there exists 24 PETE L. CLARK a unique R-module map f : M → N such that fi = f ◦ ιi for all i ∈ I . Thus M satisfies the universal mapping property of the direct sum. As a matter of notation, for n ∈ Z+ , Rn := ⊕n i=1 R, R0 = 0. Exercise X.X: Work out the analogue of Exercise X.X for direct products. Exercise X.X: a) Suppose that M is an R-module and S is a subset of M . Show that the intersection of all R-submodules of M containing S is an R-submodule, and is contained in every R-submodule that contains S . We call it the R-submodule generated by S . b) ∑ If S = {si }i∈I , show that the R-module generated by S is the set of all sums i∈J ri si , where J is a finite subset of S . Exercise X.X: Suppose that k is a field. Show that the terms “k -module” and “vector space over k ” are synonymous. One can therefore view the theory of R-modules as a generalization of vector spaces to arbitrary rings. But really this is something like a zeroth order approximation of the truth: for a general ring R, the theory of R-modules is incomparably richer than the theory of vector spaces over a field. There are two explanations for this. First, even when working with very simple R-modules such as Rn , the usual linear algebra notions of linear independence, span and basis remain meaningful, but behave in unfamiliar ways: Call a subset S of an R-module M linearly independent if for every finite subset m1 , . . . , mn of S and any r1 , . . . , rn ∈ R, r1 m1 + . . . + rn mn = 0 implies r1 = . . . = rn = 0. Say that S spans R if the R-submodule generated by S is R, and finally a basis for an R-module is a subset which is both linearly independent ⊕ and spanning. For example, for any set I , the R-module i R has a basis ei . In linear algebra – i.e., when R is a field – every R-module has a basis.8 However the situation is quite different over a general ring: Theorem 11. a) Let M be an R-module. Suppose that S ⊂ R is a basis. Then M ⊕ is isomorphic as an R-module to s∈S R. ⊕ b) Let S ⊕ any set, and consider the R-module RS := s∈S R. For each s ∈ S , be let es ∈ s∈S R be the element whose s-coordinate is 1 and all of whose other coordinates are 0. Then set {es }s∈S is a basis for RS . Exercise X.X: Prove Theorem X.X. A module which has a basis – so, by the theorem, admits an isomorphism to for some index set S – is called free. ⊕ s∈S R Exercise X.X: Show that a nonzero free R-module is faithful. 8This uses, and is in fact equivalent to, the Axiom of Choice, but the special case that any vector space with a finite spanning set has a basis does not. COMMUTATIVE ALGEBRA 25 Let us examine the case of modules over R = Z, i.e., of abelian groups. In this case, the term free abelian group is synonmyous with free Z-module. Probably the reader knows that not all abelian groups are free, but in case not: for any integer n > 1, the Z-module Z/nZ is not free, since it has nonzero annihilator nZ. Thus Z/nZ does not have a basis as a Z-module, and indeed has no nonempty linearly independent subsets! In fact: Proposition 12. For a commutative ring R, TFAE: (i) Every R-module is free. (ii) R is a field. Proof. As discussed above, (ii) =⇒ (i) is a fundamental theorem of linear algebra, so we need only concern ourselves with the converse. But if R is not a field, then (Exercise X.X) there exists a nonzero proper ideal I , and then R/I is a nontrivial R-module with 0 ̸= I = ann(R/I ), so by Exercise X.X R/I is not free. Noncommutative Remark: If R is a non-commutative ring such that every left Rmodule is free, then the above argument shows R has no nonzero proper twosided ideals, so is what is called a simple ring. But a noncommutative simple ring may still admit a nonfree module. For instance, let k be a field and take R = M2 (k ), the 2 × 2 matrix ring over k . Then k ⊕ k is a left R-module which is not free. However, suppose R is a ring with no proper nontrivial one-sided ideals. Then R is a division ring – i.e., every nonzero element of R is a unit – and every R-module is free. In linear algebra – i.e., when R is a field – every linearly independent subset of an R-module can be extended to a basis. Over a general ring this does not hold even for free R-modules. For instance, take R = M = Z. A moment’s thought reveals that the only two bases are {1} and {−1}, whereas the linearly independent sets are precisely the singleton sets {n} as n ranges over the nonzero integers. Final Remark: Note well the form of Proposition 12: we assume that R is a commutative ring for which all R-modules (or, more generally, some nice class of R-modules) satisfy some nice property, and we deduce a result on the structure of R. Such problems – roughly, “inverse problems” – have a broad aesthetic appeal throughout mathematics, and provide one of the major motivations for studying modules above and beyond their linear-algebraic origins. We will see other such characterizations later on. 3.2. Finitely presented modules. One of the major differences between abelian groups and nonabelian groups is that a subgroup N of a finitely generated abelian group M remains finitely generated, and indeed, the minimal number of generators of the subgroup N cannot exceed the minimal number of generators of M , whereas this is not true for nonabelian groups: e.g. the free group of rank 2 has as subgroups free groups of every rank 0 ≤ r ≤ ℵ0 . (For instance, the commutator subgroup is not finitely generated.) Since an abelian group is a Z-module and every R-module has an underlying abelian group structure, one might well expect the situation for R-modules to be similar to 26 PETE L. CLARK that of abelian groups. We will see later that this is true in many but not all cases: an R-module is called Noetherian if all of its submodules are finitely generated. Certainly a Noetherian module is itself finitely generated. The basic fact here – which we will not prove until §X.X , despite its relative simplicity – is a partial converse: if the ring R is Noetherian, any finitely generated R-module is Noetherian. Note that we can already see that the Noetherianity of R is necessary: if R is not Noetherian, then by definition there exists an ideal I of R which is not finitely generated, and this is nothing else than a non-finitely generated R-submodule of R (which is itself generated by the single element 1.) Thus the aforementioned fact about subgroups of finitely generated abelian groups being finitely generated holds because Z is a Noetherian ring. When R is not Noetherian, it becomes necessary to impose stronger conditions than finite generation on modules. One such condition indeed comes from group theory: recall that a group G is finitely presented if it is isomorphic to the quotient of a finitely generated free group F by the least normal subgroup N generated by a finite subset x1 , . . . , xm of F . Proposition 13. For a finitely generated R-module M , TFAE: (i) There exist non-negative integers m, n and an exact sequence R m → R n → M → 0. (ii) M is the quotient of an f.g. free R-module Rn by some f.g. submodule N . A module M satisfying these equivalent conditions is said to be finitely presented. Proof. That (i) implies (ii) is immediate. Conversely, let M = Rn /N where N is finitely generated. Then there exists a surjection Rm → N and thus the sequence Rm → Rn → M → 0 is exact. Proposition 14. Let ψ ϕ 0→K→N →M →0 be a short exact sequence of R-modules, with M finitely presented and N finitely generated. Then K is finitely generated. Proof. (Matsumura) By definition of finitely presented, we can place M in an exact sequence (1) f Rm → Rn M → 0 for some m, n ∈ N. For 1 ≤ i ≤ n, let ei be the ith standard basis element of M , let mi = f (ei ) be the image in M , and choose ni ∈ N any element in ϕ−1 (mi ). Then there is a unique R-module homomorphism α : Rn → N given by α(ei ) = ni , which restricts to an R-module homomorphism β : B m → K . Altogether we get a commutative diagram f Rm −→ Rn −→ M −→ 0 ψ ϕ 0 −→ K −→ N −→ M. COMMUTATIVE ALGEBRA 27 The rest of the proof is essentially a diagram chase. Suppose N = ⟨ξ1 , . . . , ξk ⟩R , and choose v1 , . . . , vk ∈ Rn such that ϕ(ξi ) = f (vi ). Put ′ ξi = ξi − α(vi ). ′ Then φ(ξi ) = 0, so there exist unique ηi ∈ K such that ′ ξi = ψ (ηi ). We claim that K is generated as an R-module by β (Rm ) and η1 , . . . , ηk and thus is finitely generated. Indeed, for η ∈ K , there are r1 , . . . , rk ∈ R such that ∑ ψ (η ) = ri ξi . i Then ψ (η − ∑ ri ηi ) = i Since we may write ∑ ∑ ′ ri (ξi − ξi ) = α( ri vi ). i i ∑ ∑ 0 = ϕ(α( ri vi )) = f ( ri vi ), i ∑ i = g (u) with u ∈ Rm . Then ∑ ∑ ψ (β (u)) = α(g (u)) = α( ri vi ) = ψ (η − ri ηi ). i ri vi i Since ψ is injective, we conclude η = β ( u) + i ∑ ri ηi . i Exercise X.X: Let 0 → M ′ → M → M ′′ → 0 be a short exact sequence of R-modules. a) Show that if M ′ and M ′′ are both finitely presented, so is M . b) Show that if M is finitely presented and M ′ is finitely generated, then M ′′ is finitely presented. A stronger condition yet is the following: an R-module M is coherent if it is finitely generated and every finitely generated submodule is finitely presented. Evidently coherent implies finitely presented implies finitely generated, and all three coincide over a Noetherian ring. The significance of coherence lies in the following: Theorem 15. Let R be a ring (not necessarily commutative, but with unity). a) The category of all left R-modules is an abelian category. b) The category of all coherent left R-modules is an abelian category. c) In particular, if R is left Noetherian, the category of all finitely generated left R-modules is an abelian category. d) There exists a commutative ring R for which the category of all finitely generated (left) R-modules is not abelian. The proof of this result – and even an explanation of the term “abelian category” is beyond the scope of these notes, so this is definitely an ancillary remark. Nevertheless we hope that it will be of some use to students of algebraic geometry: for instance, it explains the fact that Hartshorne only defines coherent sheaves of 28 PETE L. CLARK modules on a scheme X in the case that the scheme is Noetherian and suggests the correct (and more subtle) definition in the non-Noetherian case. 3.3. Torsion and torsionfree modules. Let R be a domain, and let M be an R-module. An element x ∈ M is said to be torsion if there exists 0 ̸= a ∈ R such that ax = 0. Equivalently, the annihilator ann(x) = {a ∈ R | ax = 0} is a nonzero ideal of R. We define M [tors] to be the set of all torsion elements of M . It is immediate to see that M [tors] is a submodule of M . We say that M is a torsion R-module if M = M [tors] and that M is torsionfree if M [tors] = 0. Exercise: Let 0 → M1 → M → M2 → 0 be an exact sequence of modules. a) Show that if M is torsion, so are M1 and M2 . b) If M1 and M2 are torsion modules, must M be torsion? c) Show that if M is torsionfree, show that so is M1 , but M2 need not be. d) If M1 and M2 are torsionfree, must M be torsionfree? Proposition 16. Let R be an integral domain and M an R-module. a) The quotient M/M [tors] is torsionfree. b) If M is finitely generated and torsionfree, then it is isomorphic to a submodule of a finitely generated free module. Proof. a) Put N = M/M [tors], and let x ∈ N be such that there exists 0 ̸= a ∈ R with ax = 0. Let x be any lift of x to M ; then there exists t ∈ M [tors] such that ax = t. By definition of torsion, there exists a′ ∈ R such that a′ t = 0, so a′ ax = a′ t = 0. Since R is a domain, a′ a is nonzero, so x ∈ M [tors] and x = 0. c) Suppose M = ⟨x1 , . . . , xr ⟩. We may of course assume that M ̸= 0 and hence that r ≥ 1 and all the xi are nonzero. Further, after reordering the xi ’s if necessary, there exists a unique s, 1 ≤ s ≤ r, such that {x1 , . . . , xs } is linearly independent over R but for all i with s < i ≤ r, {x1 , . . . xs , xi } is linearly dependent over R. Then F = ⟨x1 , . . . , xs ⟩ ∼ Rs , so we are done if s = r. If s < ∏ then for each r, = i > s there exists 0 ̸= ai ∈ R such that ai xi ∈ F . Put a = s<i≤r ai : then aM ⊂ F . Let [a] : M → M denote multiplication by a. Since M is torsionfree, [a] is injective hence gives an R-module isomorphism from M to a submodule of the finitely generated free module F . Exercise: Show that the torsionfree Z-module (Q, +) is not isomorphic to a submodule of any finitely generated free Z-module. Thus – even for very nice rings! – the hypothesis of finite generation is necessary in Proposition 16. 3.4. Tensor and Hom. 3.4.1. Tensor products. We assume that the reader has some prior familiarity with tensor products, say of vector spaces and/or of abelian groups. The first is an instance of tensor products of k -modules, for some field k , and the second is an instance of tensor products of Z-modules. We want to give a general definition of M ⊗R N , where M and N COMMUTATIVE ALGEBRA 29 are two R-modules. There are two ways to view the tensor product construction: as a solution to a universal mapping problem, and as a generators and relations construction. They are quite complementary, so it is a matter of taste as to which one takes as “the” definition. So we will follow our taste by introducing the mapping problem first: Suppose M , N , P are R-modules. By an R-bilinear map f : M × N → P we mean a function which is separately R-linear in each variable: for all m ∈ M , the mapping n → f (m, n) is R-linear, and for each n ∈ N , the mapping m → f (m, n) is R-linear. Now consider all pairs (T, ι), where T is an R-module and ι : M × N → T is an R-bilinear map. A morphism from (T, ι) to (T ′ , ι′ ) will be an R-module homomorphism h : T → T ′ such that ι′ = h ◦ ι. By definition, a tensor product M ⊗R N is an initial object in this category: i.e., it comes equipped with an R-bilinear map M × N raM ⊗R N such that any R-bilinear map f : M × N → P factors through it. As usual, the initial object of a category is unique up to unique isomorphism provided it exists. As for the existence, we fall back on the generators and relations construction. Namely, we begin with the free R-module F whose basis is M × N , and we write the basis elements (purely formally) as m ⊗ n. We then take the quotient by the submodule generated by the following relations R: (x + x′ ) ⊗ y − x ⊗ y − x′ ⊗ y, x ⊗ (y + y ′ ) − x ⊗ y − x ⊗ y ′ , (ax) ⊗ y − a(x ⊗ y ), x ⊗ (ay ) − a(x ⊗ y ). It is then easy to see that the quotient map M × N → F/N satisfies all the properties of a tensor product (details left to the reader). Note that the general element of M ⊗R N is not a single element of the form x ⊗ y but rather a finite sum of such elements. (Indeed, from the free R-module, every element can be represented by a finite R-linear combination of elements of the form x ⊗ y , but the last two defining relations in the tensor product allow us to change ri (x ⊗ y ) to either (ri x) ⊗ y or x ⊗ (ri y ).) Of course, this representation of an element of the tensor product need not be (and will never be, except in trivial cases) unique. One can also take the tensor product of R-algebras: if R is a (commutative!) ring and A and B are commutative R-algebras, then on the tensor product A ⊗R B we have a naturally defined product, induced by (a1 ⊗ b1 ) · (a2 ⊗ b2 ) := (a1 a2 ⊗ b1 b2 ). We have to check that this is well-defined, a task which we leave to the reader (or see [AM, pp. 30-31])). The tensor product of algebras is a powerful tool – e.g. in the structure theory of finite-dimensional algebras over a field, or in the theory of linear disjointness of field extensions – and is given misleadingly short shrift in most elementary treatments. Base change: Suppose that M is an R-module and f : R → S is a ring homomorphism. Then S is in particular an R-module, so that we can form the tensor product S ⊗R M . This is still an R-module, but it is also an S -module in an evident 30 PETE L. CLARK ∑ ∑ way: s • ( i si ⊗ mi ) := i ssi ⊗ mi . This is process is variously called scalar extension, base extension or base change. Note that this process is functorial, in the following sense: if f : M → M ′ is an R-algebra homomorphism, then there exists an induced S -algebra homomorphism S ⊗R M → S ⊗R M ′ , given by s ⊗ m → s ⊗ f (m). Exercise X.X: If M is a finitely generated R-module and f : R → S is a ring homomorphism, then S ⊗R M is a finitely generated S -module. Exercise X.X: Let A and B be rings, M an A-module, P a B -module, and N an (A, B )-bimodule. Then M ⊗A N is naturally a B -module, N ⊗B P is naturally an A-module, and (M ⊗A N ) ⊗B P ∼ M ⊗A (N ⊗B P ). = Exercise X.X: Let R be a commutative ring, I and ideal of R and M an R-module. a) Show that there is a well-defined R-bilinear map R/I × M → M/IM given by (r + I, m) → rm + I . Thus there is an induced homomorphism of R-modules φ : R/I ⊗R M → M/IM. b) Show that φ is an isomorphism of R-modules. Proposition 17. Let R be a commutative ring, M an R-module and {Ni }i∈I a directed system of R-modules. Then the R-modules lim(M ⊗ Ni ) and M ⊗ (lim Ni ) − → − → are canonically isomorphic. Exercise X.X: Prove Proposition 17. 3.4.2. Hom modules. 3.4.3. Compatibilities between ⊗ and Hom. 3.5. Projective modules. 3.5.1. Basic equivalences. Proposition 18. For an R-module P , TFAE: (i) There exists an R-module Q such that P ⊕ Q is a free R-module. (ii) If π : M → N is a surjective R-module homomorphism and φ : P → N is a homomorphism, then there exists at least one R-module homomorphism Φ : P → M such that φ = π ◦ Φ. (iii) If π : M → N is a surjection, then the natural map Hom(P, M ) → Hom(P, N ) given by Φ → π ◦ Φ is surjective. (iv) The functor Hom(P, ) is exact. (v) Any short exact sequence of R-modules q 0→N →M →P →0 splits: there exists an R-module map σ : P → M such that q ◦ σ = 1P and thus an internal direct sum decomposition M = N ⊕ σ (P ). A module satisfying these equivalent conditions is called projective. COMMUTATIVE ALGEBRA 31 Proof. (i) =⇒ (ii): Let F ∼ P ⊕ Q be a free module. Let {fi } be a free basis for = F and let {pi } be the corresponding generating set for P , where pi is the image of fi under the natural projection P ⊕ Q → P . Put ni = φ(pi ). By surjectivity of π , let mi ∈ π −1 (ni ). By the freeness of F , there is a unique R-module homomorphism h : F → M carrying each fi to mi . Pull h back to P via the natural inclusion P → F . Then h : P → M is such that π ◦ f = φ. (ii) =⇒ (i): As for any R-module, there exists a free R-module F and a surjection π : F → P . Applying (ii) with N = p and φ : P → N the identity map, we get a homomorphism Φ : P → F such that π ◦ φ = 1P . It follows that F = Φ(P ) ⊕ ker(π ) is an internal direct sum decomposition. (ii) ⇐⇒ (iii): (iii) is nothing more than a restatement of (ii), as we leave it to the reader to check. (iii) ⇐⇒ (iv): To spell out (iv), it says: if 0 → M ′ → M → M ′′ → 0 is a short exact sequence of R-modules, then the corresponding sequence 0 → Hom(P, M ′ ) → Hom(P, M ) → Hom(P, M ′′ ) → 0 is exact. It is easy to see that for any R-module P , the sequence 0 → Hom(P, M ′ ) → Hom(P, M ) → Hom(P, M ′′ ) is exact – i.e., Hom(P, ) is left exact – so (iv) amounts to: for any surjection M → M ′′ , the corresponding map Hom(P, M ) → Hom(P, M ′′ ) is surjective, and this is condition (iii). (ii) =⇒ (v): Given q 0 → N → M → P → 0, we apply (ii) to the identity map 1P : P → P and the surjection q : M → P , getting a map σ : P → M such that q ◦ σ = 1P , so σ is a section as required. (v) =⇒ (i): Choosing a set of generators for P gives rise to a surjective homomorphism q : F → P from a free R-module F to P and thus a short exact sequence q 0 → Ker q → F → P → 0. By hypothesis, there exists a section σ : P → F and thus an internal direct sum decomposition F ∼ Ker(q ) ⊕ σ (P ) ∼ Ker(q ) ⊕ P . = = Exercise X.X: Give a direct proof that (v) =⇒ (ii) in Proposition 18. (Suggestion: Given the surjection q : M → N and the map π : P → N , form the short exact sequence 0 → K → M → N → 0 and show that it is mapped to by a short exact sequence 0 → K → M ×N P → P → 0, where M ×N P = {(x, y ) ∈ M × P | q (x) = π (y )} is the fiber product of M and P over N .) Exercise X.X: Use Proposition 18 to show, several times over, that a free R-module is projective. Exercise X.X: Let {Mi }i∈I be an index family of R-modules. Show that the direct ⊕ sum M = i∈I Mi is projective iff each Mi is projective. 32 PETE L. CLARK Exercise X.X: a) Show that the tensor product of two free R-modules is free. b) Show that the tensor product of two projective R-modules is projective. Exercise X.X: Show that a finitely generated projective module is finitely presented. (Hint: the problem is that over a not-necessarily-Noetherian ring, a submodule of a finitely generated module need not be finitely generated. However, a direct summand of a finitely generated module is always finitely generated: why?) 3.5.2. Linear algebraic characterization of projective modules. Let R be a commutative ring, n ∈ Z+ , and let P be an element of the (noncommutative!) ring Mn (R) of n × n matrices with entries in R such that P 2 = P . There are several times for such a matrix. The pure algebraist would call such a matrix idempotent, for that is the name of an element in any ring which is equal to its square. A geometrically minded algebraist however may call such a matrix a projection, the idea being that the corresponding R-module endomorphism of Rn “projects” Rn onto the submodule P (Rn ). Proposition 19. An R-module M is finitely generated and projective iff it is, up to isomorphism, the image of a projection: i.e., iff there exists n ∈ Z+ and a matrix P ∈ Mn (R) with P = P 2 such that M ∼ P (Rn ). = Proof. Suppose first that M is a finitely generated projective R-module. Since M is finitely generated,, there exists n ∈ Z+ and a surjective R-module homorphism π : Rn → M . Since M is projective, this homomorphism has a section σ : M → Rn , and we may thus write Rn = σ (M ) ⊕ M ′ . Put P = σ ◦ π ∈ EndR (Rn ). Then P (Rn ) = σ (π (Rn )) = σ (M ) ∼ M and = P 2 = σ ◦ (π ◦ σ ) ◦ π = σ ◦ 1M ◦ π = σ ◦ π = P. Conversely, suppose that there exists P ∈ EndR (Rn ) with P 2 = P and let M ∼ = P (Rn ). Then – since P (1 − P ) = 0 – Rn = P (Rn ) ⊕ (1 − P )(Rn ), exhibiting P (Rn ) as a direct summand of a free module.9 3.5.3. The Dual Basis Lemma. Proposition 20. (Dual Basis Lemma) For an R-module M , TFAE: (i) There exists an index set I , elements {ai }i∈I of M and homomorphisms {fi : M → R}i∈I such that for each a ∈ M , {i ∈ I | fi (a) ̸= 0} is finite, and ∑ a= fi (a)ai . i ∈I (ii) M is projective. Proof. (i) =⇒ (ii): Let F be the free R-module with basis elements {ei }i∈I , and define f : F → M by f (ei ) = ai . Then the map ι : M → F given by ∑ ι(a) = i∈I fi (a)ei is a section of f , so M is a direct summand of F . ⊕ (ii) =⇒ (i): Let f : F = i∈I R → M be an epimorphism from a free R-module 9Note that this part of the proof redeems the pure algebraist: this the decomposition afforded by the pair of orthogonal idempotents P, 1 − P . COMMUTATIVE ALGEBRA 33 onto M . Since M is projective, there exists a section ι : M → F . If {ei }i∈I is the standard basis of F , then for all a ∈ M , the expression ∑ ι(a) = fi (a)ei i ∈I defines the necessary family of functions fi : M → R. Exercise: Let P be a projective R-module. Show that one can can find a finite index set I satisfying condition (i) of the Dual Basis Lemma iff P is finitely generated. 3.5.4. Projective versus free. Having established some basic facts about projective modules, we should now seek examples in nature: which modules are projective? Note that by Exercise X.X any free module is projective. But this surely counts as a not very interesting example! Indeed the following turns out to be one of the deepest questions of the subject. Question 1. When is a projective module free? We want to give examples to show that the answer to Question 1 is not “always”. But even by giving examples one wades into somewhat deep waters. The following is the one truly “easy” example of a non-free projective module I know. Example: Suppose R1 and R2 are nontrivial rings. Then the product R = R1 × R2 admits nonfree projective modules. Indeed, let P be the ideal R1 × {0} and Q the ideal {0} × R2 . Since R = P ⊕ Q, P and Q are projective. On the other hand P cannot be free because taking e := (0, 1) ∈ R, we have eP = 0, whereas eF ̸= 0 for any nonzero free R-module F (and of course, Q is not free either for similar reasons). One way to construe Question 1 is to ask for the class of rings over which every projective module is free, or over which every finitely generated projective module is free. I actually do not myself know a complete answer to this question, but there are many interesting and important special cases. Recall the following result from undergraduate algebra. Theorem 21. A finitely generated module over a PID is free iff it is torsionfree. Of course submodules of torsionfree modules are torsionfree, so projective implies torsionfree. We deduce: Corollary 22. A finitely generated projective module over a PID is free. Theorem 21 does not extend to all torsionfree modules: for instance, the Z-module Q is torsionfree but not free. However Corollary 22 does extend to all modules over a PID. The proof requires transfinite methods and is given in §3.10. Recall that a ring R is local if it has a unique maximal ideal. It is convenient to reserve the notation m for the unique maximal ideal of a local ring and speak of “the local ring (R, m)”. We want to show that every finitely generated projective module over a local ring is free. First a few preliminaries. 34 PETE L. CLARK Let f : R → S be a homomorphism of rings. Then necessarily f induces a homomorphism f × : R× → S × on unit groups: if xy = 1, then f (x)f (y ) = f (1) = 1, so units get mapped to units. But what about the converse: if x ∈ R is such that f (x) is a unit in S , must x be a unit in R? It’s a nice idea, but it’s easy to see that this need not be the case. For instance, let a > 1 be any positive integer. Then a is not a unit of Z, but for each prime p > a, the image of a in the quotient ring Z/pZ is a unit. Too bad! Let us not give up so soon: a conjecture may fail, but a definition cannot: say a homomorphism f : R → S of rings is unit-faithful if for all x ∈ R, f (x) ∈ S × =⇒ x ∈ R× . Lemma 23. If (R, m) is a local ring, the quotient map q : R → R/m is unit-faithful. Proof. An element of any ring is a unit iff it is contained in no maximal ideal, so in a local ring we have R× = R \ m. Moreover, since m is maximal, R/m is a field. Thus, for x ∈ R, q (x) ∈ (R/m)× ⇐⇒ x ∈ m ⇐⇒ x ∈ R× . / Theorem 24. A finitely generated projective module over a local ring is free. Proof. Let P be a finitely generated projective module over the local ring (R, m). We may find Q and n ∈ Z+ such that P ⊕ Q = Rn . Now tensor with R/m: we get a direct sum decomposition P/mP ⊕ Q/mQ = (R/m)n . Since R/m is a field, all R/m-modules are free. Choose bases {pi } for P/mP and {qj } for Q/mQ, and for all i, j , lift each pi to an element pi of P and each qj to an element qj of Q. Consider the n × n matrix A with coefficients in R whose columns are p1 , . . . , pa , q1 , . . . , qb . The reduction modulo m of A is a matrix over the field R/m whose columns form a basis for (R/m)n , so its determinant is a unit in (R/m)× . Since det(M (mod m)) = det(M ) (mod m), Lemma 23 implies that det(M ) ∈ R× , i.e., M is invertible. But this means that its columns are linearly independent, so p1 , . . . , pa , a priori only a generating set for the R-module P , is in fact a basis. Once again, in Section 3.9 this result wll be improved upon: it is a celebrated theorem of Kaplansky that any projective module over a local ring is free. Finally, we give without proof a celebrated result of H. Bass with a clear moral: finitely generated projective modules are usually much more interesting than infinitely generated projective modules. Here let us be sure that by an infinitely generated R-module, we mean an R-module which is not finitely generated.10 Theorem 25. (Bass [Bas63]) Let R be connected (i.e., without nontrivial idempotents) Noetherian ring. Then any infinitely generated projective R-module is free. Much more interesting is an example of a finitely generated projective, nonfree module over an integral domain. Probably the first such examples come from nonprincipal ideals in rings of integers of number fields with class number greater than 1. To give such an example with proof of its projectivity this early in the day, we 10A priori it would be reasonable to take “infinitely generated R-module” to mean a module which possesses an infinite generating set, but a moment’s thought shows that an R-module has this property iff it is infinite, so it is more useful to define “infinitely generated” as we have. COMMUTATIVE ALGEBRA 35 require a little preparation.11 Two ideals I and J in a ring R are comaximal if I + J = R. More generally, a family {Ii } of ideals in a ring is pairwise comaximal if for all i ̸= j , I + J = R. Lemma 26. Let I , J , K1 , . . . , Kn be ideals in the ring R. Put K = K1 · · · Kn . a) We have (I + J )(I ∩ J ) ⊂ IJ . b) If I and J are comaximal, IJ = I ∩ J . Proof. a) (I + J )(I ∩ J ) = I (I ∩ J ) + J (I ∩ J ) ⊂ IJ + IJ = IJ . b) If I + J = R, the identity of part a) becomes I ∩ J ⊂ IJ . Since the converse inclusion is valid for all I and J , the conclusion follows. Proposition 27. Let I and J be comaximal ideals in a domain R, and consider the R-module map q : I ⊕ J → R given by (x, y ) → x + y . Then: a) The map q is surjective. b) Ker(q ) = {(x, −x) | x ∈ I ∩ J }, hence is isomorphic as an R-module to I ∩ J . c) We have an isomorphism of R-modules I ⊕ J ∼ IJ ⊕ R. = d) Thus if IJ is a principal ideal, I and J are projective modules. Proof. It is clear that for any ideals I and J , the image of the map q is the ideal I + J , and we are assuming I + J = R, whence part a). Part b) is essentially immediate: details are left to the reader. Combining parts a) and b) we get a short exact sequence 0 → I ∩ J → I ⊕ J → R → 0. But R is free, hence projective, and thus the sequence splits, giving part c). Finally, a nonzero principal ideal (x) in a domain R is isomorphic as an R-module to R itself: indeed, multiplication by x gives the isomorphism R → (x). So if IJ is principal, I ⊕ J ∼ R2 and I and J are both direct summands of a free module. = In particular, if we can find in a domain R two comaximal nonprincipal ideals I and J with IJ principal, then I and J are finitely generated projective nonfree R-modules. The following exercise asks you to work through an explicit example. √ Exercise X.X: Let R = Z[ −5], and put √ √ p1 = ⟨3, 1 + −5⟩, p2 = ⟨3, 1 − −5⟩. a) Show that R/p1 ∼ R/p2 ∼ Z/3Z, so p1 and p2 are maximal ideals of R. = = b) Show that p1 + p2 = R (or equivalently, that p1 ̸= p2 ). c) Show that p1 p2 = (3). d) Show that neither p1 nor p2 is principal. √ √ (Suggestion: show that if p1 = (x + −5y ) then p2 = (x − −5y ) and thus there are integers x, y such that x2 + 5y 2 = ±3.) e) Conclude that p1 and p2 are (in fact isomorphic) nonfree finitely generated projective modules over the domain R. 11Here we wish to acknowledge our indebtedness to K. Conrad: we took our inspiration for Proposition 27 and the following Exercise from Example 3.1 of http://www.math.uconn.edu/∼kconrad/blurbs/linmultialg/splittingmodules.pdf. 36 PETE L. CLARK f) Show that p2 is principal, and thus that the class of p in K0 (R) is 2-torsion. This construction looks very specific, and the number-theoretically inclined reader is warmly invited to play around with other quadratic rings and more general rings of integers of number fields to try to figure out what is really going on. From our perspective, we will (much later on) gain a deeper understanding of this in terms of the concepts of invertible ideals, the Picard group and Dedekind domains. Example: Let X be a compact space, and let C (X ) be the ring of continuous real-valued functions on X . The basic structure of these rings is studied in §X.X . Let E → X be a real topological vector bundle over X . Then the group Γ(E ) of global sections is naturally a module over C (X ). In fact it is a finitely generated projective module, and all finitely generated projective C (X )-modules arise faithfully in this way: the global section functor gives a categorical equivalence between vector bundles on X and finitely generated projective modules over C (X ). This is a celebrated theorem of R.G. Swan, and Section X is devoted to giving a self-contained discussion of it, starting from the definition of a vector bundle. In particular, via Swan’s Theorem basic results on the tangent bundles of compact manifolds translate into examples of finitely generated projective modules: for instance, an Euler characteristic argument shows that the tangent bundle of any even-dimensional sphere S 2k is nontrivial, and thus Γ(T S 2k ) is a finitely generated nonfree C (S 2k )-module! Following Swan, we will show that examples of nonfree projective modules over more traditional rings like finitely generated R-algebras follow from examples like these. Example: Let k be a field and R = k [t1 , . . . , tn ] be the polynomial ring over k in n indeterminates. When n = 1, R is a PID, so indeed every finitely generated R-module is projective. For n > 1, the situation is much less clear, but the problem of freeness of finitely generated projective R-modules can be stated geometrically as follows: is any algebraic vector bundle on affine n-space An algebraically trivial? /k When k = C, the space AnC = Cn in its usual, Euclidean topology is contractible, / which by basic topology implies that any continuous C-vector bundle on An is (continuously) trivial. Moreover, relatively classical complex variable theory shows that any holomorphic vector bundle on An is (holomorphically) trivial. But asking the transition functions and the trivialization to be algebraic – i.e., polynomial functions – is a much more stringent problem. In his landmark 1955 paper FAC, J.-P. Serre noted that this natural problem remained open for algebraic vector bundles: he was able to prove only the weaker result that a finitely generated projective R-module M is stably free – i.e., there exists a finitely generated free module M such that M ⊕ F is free. This became known as Serre’s Conjecture (to his dismay) and was finally resolved independently in 1976 by D. Quillen [Qui76] and A. Suslin [Su76]: indeed, every finitely generated projective R-module is free. Quillen received the Fields Medal in 1978. Fields Medals are not awarded for the solution of any single problem, but the prize committee writes an official document describing the work of each winner that they found particularly meritorious. In this case, it was made clear that Quillen’s resolution of Serre’s Conjecture was one of the reasons he received the prize. All this for modules over a polynomial ring! COMMUTATIVE ALGEBRA 37 For more information on Serre’s Conjecture, the reader could do no better than to consult a recent book of T.Y. Lam [Lam06]. Exercise (K0 (R)): From a commutative ring R, we will construct another commutative ring K0 (R) whose elements correspond to formal differences of finite rank projective modules. More precisely: a) Let M0 (R) denote the set of all isomorphism classes of finitely generated projective modules. For finitely generated projective modules P and Q we define [P ] + [Q] = [P ⊕ Q], [P ] · [Q] = [P ⊗ Q]. Check that this construction is well-defined on isomorphism classes and endows M0 (R) with the structure of a commutative semiring with unity. What are the additive and mulitplicative identity elements? b) Define K0 (R) as the Grothendieck group of M0 (R), i.e., as the group completion of the commutative monoid M0 (R). Convince yourself that K0 (R) has the structure of a semiring. The elements are of the form [P ] − [Q], and we have [P1 ] − [Q1 ] = [P2 ] − [Q2 ] ⇐⇒ there exists a finitely generated projective R-module M with ∼ P1 ⊕ Q2 ⊕ M = P2 ⊕ Q1 ⊕ M. In particular, if P and Q are projective modules, then [P ] = [Q] in K0 (R) iff [P ] and [Q] are stably isomorphic, i.e., iff they become isomorphic after taking the direct sum with some other finitely generated projective module M . c) Show that we also have [P ] = [Q] iff there exists a nitely generated free module Rn such that P ⊕ Rn ∼ Q ⊕ Rn . In particular, [P ] = [0] = 0 iff P is stably free: there exists a = finitely generated free module F such that P ⊕ F is free. d) Show that M0 (R) is cancellative iff every stably free finitely generated projective module is free. e)* Find a ring R admitting a finitely generated projectve module which is stably free but not free. f) Show that the mapping Rn → [Rn ] induces an injective homomorphism of rings ˜ Z → K0 (R). Define K0 (R) to be the quotient K0 (R)/Z. Show that if R is a PID ˜ 0 (R) = 0. then K The following exercise explains why in the definition of K0 (R) we restrict to finitely generated projective modules. Exercise (Eilenberg Swindle): Let us say that a projective module P is weakly stably free if there exists a not necessarily finitely generated free module F such that P ⊕ F is free. Show that every projective module is weakly stably free. (Hint: if P ⊕ Q is free, take F = P ⊕ Q ⊕ P ⊕ Q ⊕ . . ..) 3.6. Injective modules. 3.6.1. Basic equivalences. Although we will have no use for them in the sequel of these notes, in both commutative and (especially) homological algebra there is an important class of modules “dual” to the projective modules. They are characterized as follows. 38 PETE L. CLARK Proposition 28. For a module E over a ring R, the following are equivalent: (ii) If ι : M → N is an injective R-module homomorphism and φ : M → E is any homomorphism, there exists at least one extension of φ to a homomorphism Φ : N → E. (iii) If ι : M → N is an injection, the natural map Hom(N, E ) → Hom(M, E ) is injective. (iv) The (contravariant) functor Hom(E, ) is exact. (v) Any short exact sequence of R-modules ι 0→E→M →N →0 splits: there exists an R-module map π : M → E such that π ◦ ι = 1E and thus an internal direct sum decomposition M = ι(E ) ⊕ ker(π ) ∼ E ⊕ N . = A module satisfying these equivalent conditions is called injective. Exercise: Prove Proposition 28. Exercise: Show that an R-module E is injective iff whenever E is a submodule of a module M , E is a direct summand of M . Remark: Note that the set of equivalent conditions starts with (ii)! This is to facilitate direct comparison to Proposition 18 on projective modules. Indeed, one should check that each of the properties (ii) through (v) are duals of the corresponding properties for projective modules: i.e., they are obtained by reversing all arrows. The difficulty here with property (i) is that if one literally reverses the arrows in the definition of free R-module to arrive at a “cofree” R-module, one gets a definition which is unhelpfully strong: the “cofree R-module on a set X ” does not exist when #X > 1! This can be remedied by giving a more refined definition of cofree module. For the sake of curiosity, we will give it later on in the exercises, but to the best of my knowledge, cofree R-modules by any definition do not play the fundamental role that free R-modules do. Exercise X.X: Show that every module over a field is injective. Exercise X.X: Show that Z is not an injective Z-module. (Thus injectivity is the first important property of modules that is not satisfied by free modules.) Exercise X.X: Let {Mi }i∈I be any family of R-modules and put M = Show that M is injective iff Mi is injective for all i ∈ I . ∏ i ∈I Mi . Exercise: For a ring R, show TFAE: (i) R is absolutely projective: every R-module is projective. (ii) R is absolutely injective: every R-module is injective. 3.6.2. Baer’s Criterion. Theorem 29. (Baer’s Criterion [Bae40]) For a module E over a ring R, TFAE: (i) E is injective. (ii) For every ideal nonzero I of R, every R-module map φ : I → E extends to an R-module map Φ : R → E . COMMUTATIVE ALGEBRA 39 Proof. (i) =⇒ (ii): this is a special case of condition (ii) of Proposition 28: take M = I , N = R. (ii) =⇒ (i): Let M be an R-submodule of N and φ : M → E an R-module map. We need to show that φ may be extended to N . Now the set P of pairs (N ′ , φ′ ) with M ⊂ N ′ ⊂ N and φ : N ′ → E a map extending φ is nonempty and has an evident partial ordering, with respect to which the union of any chain of elements in P is again an element of P . So by Zorn’s Lemma, there is a maximal element φ′ : N ′ → E . Our task is to show that N ′ = N . Assume not, and choose x ∈ N \ N ′ . Put I = (N ′ : x) = {r ∈ R | rx ⊂ N ′ }; one checks immediately that I is an ideal of R (a generalization to modules of the colon ideal we have encountered before). Consider the composite map ·x φ I → N ′ → E; by our hypothesis, this extends to a map ψ : R → E . Now put N ′′ = ⟨N ′ , x⟩ and define12 φ′′ : N ′′ → E by φ′′ (x′ + rx) = φ′ (x′ ) + ψ (r). Thus φ′′ is an extension of φ′ to a strictly larger submodule of N than N ′ , contradicting maximality. Exercise X.X: Verify that the map φ′′ is well-defined. 3.6.3. Divisible modules. Recall that a module M over a domain R is divisible if for all r ∈ R• the endomorphism r• : M → M, x → rx, is surjective. Further, we define M to be uniquely divisible if for all r ∈ R• , the endomorphism r• : M → M is a bijection. Example: The Z-modules Q and Q/Z are divisible. Q is moreover uniquely divisible but Q/Z is not. Exercise X.X: Show that a divisible module is uniquely divisible iff it is torsionfree. Exercise X.X: a) Show that a quotient of a divisible module is divisible. b) Show that arbitrary direct sums and direct products of divisible modules are divisible. Exercise X.X: Let R be a domain with fraction field K . a) Show that K is a uniquely divisible R-module. b) Let M be any R-module. Show that the natural map M ⊗ M ⊗R K is injective iff M is torsionfree. c) Show that for any R-module M , M ⊗R K is uniquely divisible. d) Show that K/R is divisible but not uniquely divisible. Exercise X.X: 12Since N ′′ need not be the direct sum of N ′ and ⟨x⟩, one does need to check that φ′′ is well-defined; we ask the reader to do so in an exercise following the proof. 40 PETE L. CLARK a) Show that a Z-module is uniquely divisible iff it can be endowed with the (compatible) structure of a Q-module, and if so this Q-module structure is unique. b) Show that a Z-module M is a subgroup of a uniquely divisible divisible Z-module iff it is torsionfree. Proposition 30. Let R be a domain and E an R-module. a) If E is injective, it is divisible. b) If E is torsionfree and divisible, it is injective. c) If R is a PID and E is divisible, it is injective. Proof. a) Let r ∈ R• . Since r is a domain, the map r• : R → R is injective. For x ∈ E , consider the R-module homomorphism φ : R → E given by 1 → x. Since E is injective, this extends to an R-module map φ : R → E . Then Φ(1) ∈ E has the property that rΦ(1) = Φ(r1) = Φ(r) = φ(1) = x, so r• is surjective on E . b) Let I be a nonzero ideal of R and φ : I → E be an R-module map. For each a ∈ I • , there is a unique ea ∈ E such that φ(a) = aea . For b ∈ I • , we have baea = bφ(a) = φ(ba) = aφ(b) = abeb ; since E is torsionfree we conclude ea = eb = e, say. Thus we may extend φ to a map Φ : R → E by Φ(r) = re. Thus E is injective by Baer’s Criterion. c) As above it is enough to show that given a nonzero ideal I of R, every homomorphism φ : I → E extends to a homomorphism R → E . Since R is a PID, we may write I = xR for x ∈ R• . Then, as in part a), one checks that φ extends to Φ iff multiplication by x is surjective on M , which it is since M is divisible. By combining Proposition 30 with Exercise X.X, we are able to show an important special case of the desired fact that every R-module can be realized as a submodule of an injective module. Namely, if M is a torsionfree module over a domain R, then M is a submodule of the uniquely divisible – hence injective – module M ⊗R K . Exercise: Let n ∈ Z+ . a) Show that as a Z-module Z/nZ is not divisible hence not injective. b) Show that as a Z/nZ module Z/nZ is divisible iff n is a prime number. c) Show that Z/nZ is always injective as a Z/nZ-module. Exercise: Let R = Z[t] and let K be its fraction field. Show that the R-module K/R is divisible but not injective. Exercise: Let R be a domain with fraction field K . a) If R = K , (of course) all R-modules are both injective and projective. b) If R ̸= K , the only R-module which is both projective and injective is 0. 3.6.4. Enough injectives. The idea of this section is to pursue the dual version of the statement “Every R-module is a quotient of a projective module”: namely we wish to show that every R-module is a submodule of an injective module. This is a good example of a statement which remains true upon dualization but becomes more elaborate to show. The projective version is almost obvious: indeed, we have the stronger result that every module is a quotient of a free module, and – as we have seen – to realize M as a quotent of a free R-module is equivalent to simply choosing a set of COMMUTATIVE ALGEBRA 41 generators for M . (But again, if we choose the most obvious definition of “cofree”, then this statement will be false.) Let k be a ring, R a k -algebra, M an R-module and N a k -module. Consider the commutative group Homk (M, N ). We may endow it with the structure of an R-module as follows: for r ∈ R and f ∈ HomZ (M, N ), (rf )(x) := f (rx). Consider the special case k = Z and N = Q/Z of the above construction. It gives HomZ (M, Q/Z) the structure of an R-module, which we denote by M ∗ and call the Pontrjagin dual of M .13 Because Q/Z is an injective Z-module, the (contravariant) functor M → M ∗ – or in other words HomZ ( , Q/Z) – is exact.14 In particular, if f : M → N is an R-module map, then f injective implies f ∗ surjective and f surjective implies f ∗ injective. As is often the case for “duals”, we have a natural map M → M ∗∗ : namely x → (f → f (x)). Lemma 31. For any R-module M , the natural map ΨM : M → M ∗∗ is injective. Proof. Seeking a contradiction, let x ∈ M • be such that Ψ(x) = 0. Unpacking the definition, this means that for all f ∈ HomZ (M, Q/Z), f (x) = 0. But since Q/Z is an injective Z-module, it suffices to find a nontrivial homomorphism Zx → Q/Z, 1 and this is easy: if x has finite order n > 1, we may map x to n , whereas if x has infinite order we may map it to any nonzero element of Q/Z. Lemma 32. Every Z-module M can be embedded into an injective Z-module. ⊕ Proof. Let I be a set of generators for the Z-module M ∗ and put F = i∈I Z, and consider the induced surjection of Z modules π : F → M ∗ . Taking ⊕ Pontrjagin ⊕ duals, we get an injection π ∗ : M ∗∗ → F ∗ = Hom( i∈I Z, Q/Z) = i∈I Q/Z. Therefore we get an injection ⊕ π ∗ ◦ ΨM : M → Q/Z. i ∈I ⊕ The Z-module i∈I Q/Z is a direct sum of divisible modules, hence divisible, hence – since Z is a PID – injective. We are done! Lemma 33. (Injective Production Lemma) Let R be a k -algebra, E an injective k -module and F a free R-module. Then Homk (F, E ) is an injective R-module. Proof. We will show that the functor HomR ( , Homk (F, E )) is exact. For any Rmodule M , the adjointness of ⊗ and Hom gives HomR (M, Homk (P, E )) = Homk (F ⊗R M, E ) so we may look at the functor M → Homk (F ⊗R M, E ) instead. This is the composition of the functor M → F ⊗R M with the functor N → Homk (N, E ). But we claim that both of these functors are exact. In the former case a moment’s thought shows this to be true. The latter case is one of our defining properties of injective modules. 13Recall that the notation M ∨ has already been taken: this is the linear dual Hom (M, R). R 14Here we are using the (obvious) fact that a sequence of R-modules is exact iff it is exact when viewed merely as a sequence of Z-modules. 42 PETE L. CLARK Remark: Soon enough we will define a flat R-module to be an R-module N such that the functor M → M ⊗R N is exact. Then Lemma 33 can be rephrased with the hypothesis that F is a at R-module, and (since as we have just seen, free R-modules are flat) this gives a somewhat more general result. Theorem 34. Every R-module can be embedded into an injective R-module. Proof. Let M be an R-module. Viewing M as a Z-module, by Lemma 32 there is an injective Z-module E1 and a monomorphism of Z-modules φ1 : M → E1 . Further, by Lemma 33, HomZ (R, E1 ) is an injective R-module. Now consider the R-module map φ : M → HomZ (R, E1 ), x → (r → φ1 (rx)). We claim that φ is a monomorphism into the injective R-module HomZ (R, E1 ). Indeed, if φ(x) = 0 then for all r ∈ R, φ1 (rx) = 0. In particular φ1 (x) = 0, so since φ1 is a monomorphism, we conclude x = 0. Exercise: Let us say that a Z-module is cofree if it is of the form F ∨ for a free Z-module F . Then the proof of Lemma 32 gives the stronger statement that every Z-module can be embedded into a cofree Z-module. Formulate a definition of cofree R-module so that the proof of Theorem 34 gives the stronger statement that every R-module can be embedded into a cofree R-module. (Hint: remember to pay attention to the difference between direct sums and direct products.) 3.6.5. Essential extensions and injective envelopes. The results of this section are all due to B. Eckmann and A. Schopf [ES53]. Proposition 35. Let M be an R-module and M ⊂R N an R-submodule. TFAE: (i) If X is any nonzero R-submodule of N , then X ∩ M is nonzero. (ii) If x ∈ N • , there exists r ∈ R such that rx ∈ M • . (iii) If φ : N → Y is an R-module map, then φ is injective iff φ|M is injective. An extension M ⊂ N satisfying these equivalent conditions is called essential. Proof. (i) =⇒ (ii): Apply (i) with X = ⟨x⟩. (ii) =⇒ (iii): Assuming (ii), let φ : N → Y be a homomorphism with φ|M is injective. It is enough to show that φ is injective. Seeking a contradiction, let x ∈ N • be such that φ(x) = 0. By (ii), there exists r ∈ R such that rx ∈ M • . But then by assumption rφ(x) = φ(rx) ̸= 0, so φ(x) ̸= 0, contradiction. (iii) =⇒ (i): We go by contraposition. Suppose there exists a nonzero submodule X of N such that X ∩ M = 0. Then the map φ : N → N/X is not an injection but its restriction to M is an injection. Proposition 36. (Tower Property of Essential Extensions) Let L ⊂ M ⊂ N be R-modules. Then L ⊂ N is an essential extension iff L ⊂ M and M ⊂ N are both essential extensions. Proof. Suppose first that L ⊂ N is an essential extension. Then for any nonzero submodule X of N , X ∩ L ̸= 0. In particular this holds for X ⊂ M , so L ⊂ M is essential. Moreover, since L ⊂ M , X ∩ L ̸= 0 implies X ∩ M ̸= 0, so M ⊂ N is essential. Conversely, suppose L ⊂ M and M ⊂ N are both essential, and let X be a nonzero submodule of N . Then X ∩ M is a nonzero submodule of M and thus (X ∩ M ) ∩ L = X ∩ L is a nonzero submodule of L. So L ⊂ N is essential. COMMUTATIVE ALGEBRA 43 So why are we talking about essential extensions when we are supposed to be talking about injective modules? The following result explains the connection. Theorem 37. For an R-module M , TFAE: (i) M is injective. (ii) M has no proper essential extensions: i.e., if M ⊂ N is an essential extension, then M = N . N . By Exercise X.X, M is a Proof. (i) =⇒ (ii): Let M be injective and M direct summand of N : there exists M ′ such that M ⊕ M ′ = N . Thus M has zero intersection with M ′ , and by criterion (ii) of Proposition 35, we must have M ′ = 0 and thus M = N . (ii) =⇒ (i): let N be an R-module such that M ⊂ N . To show that M is injective, by Exercise X.X it is equivalent to show that M is a direct summand of N . Now consider the family of submodules M ′ of N with the property that M ∩ M ′ = 0. This family is partially ordered by inclusion, nonempty, and closed under unions of chains, so by Zorn’s Lemma there exists a maximal such element M ′ . Now consider the extension M → N/M ′ : we claim it is essential. Indeed, if not, there exists x ∈ N \ M ′ such that ⟨M ′ , x⟩ ∩ M = 0, contradicting maximality of M ′ . But by hypothesis, M has no proper essential extensions: thus M = N/M ′ , i.e., M ⊕ M ′ = N and M is a direct summand of N . We say that an extension M ⊂ N is maximal essential if it is essential and there is no proper extension N ′ of N such that M ⊂ N ′ is essential. Combining Proposition 36 and Theorem 37 yields the following important result. Theorem 38. For an essential extension M ⊂ N of R-modules, TFAE: (i) M ⊂ N is maximal essential. (ii) N is injective. Exercise: To be sure you’re following along, prove Theorem 38. Once again we have a purpose in life – or at least, this subsubsection of it – we would like to show that every R-module admits a maximal essential extension and that such extensions are unique up to isomorphism over M . Moreover, a plausible strategy of proof is the following: let M be an R-module. By Theorem 34 there exists an extension M ⊂ E with E injective. Certainly this extension need not be essential, but we may seek to construct within it a maximal essential subextension N and then hope to show that M ⊂ E ′ is injective. Theorem 39. Let M be an R-module and M ⊂ E an extension with E an injective module. Let P be the set of all essential subextensions N of M ⊂ E . Then: a) P contains at least one maximal element. b) Every maximal element E ′ of P is injective. Proof. The proof of part a) is the usual Zorn’s Lemma argument: what we need to check is that the union N of any chain {Ni } of essential subextensions is again an essential subextension. Suppose for a contradiction that there exists a nonzero submodule X of N such that X ∩ M = 0. Choose x ∈ X • and put X ′ = ⟨x⟩. Then X ′ ⊂ Ni for some i and X ′ ∩ M ⊂ X ∩ M = 0, contradicting the essentialness (?!) of the extension M ⊂ Ni . Now let E ′ be a maximal essential subextension of M ⊂ E . We need to show 44 PETE L. CLARK that M ⊂ E ′ is actually a maximal essential extension: so suppose there is an essential extension E ′ ⊂ N . Let ι : M ⊂ E ′ ⊂ N be the composite map. It is a monomorphism, so by the injectivity of E the injection M ⊂ E extends to a homomorphism φ : N → E . But φ|M is an injection and M ⊂ N is an essential extension, so by condition (iii) of Proposition 35 this implies that φ itself is an injection. By maximality of E ′ among essential subextensions of M ⊂ E we must have E ′ = N . For an R-module M , we say that an extension M ⊂ E is an injective envelope (other common name: injective hull) of M if M ⊂ E is a maximal essential extension; equivalently, an essential extension with E injective. Thus Theorem 39 shows that any R-module admits an injective envelope. Proposition 40. Let R be an integral domain with fraction field K . Then R ⊂ K is an injective envelope of R. Exercise: Prove Proposition 40. (Suggestion: use the relationship between injective modules and divisible modules.) Exercise: More generally, let M be a torsionfree module over a domain R. Show that M ⊂ M ⊗R K is an injective envelope of M . Let us touch up our characterization of injective envelopes a bit. Proposition 41. (Equivalent Properties of an Injective Envelope) For an extension M ⊂ E of R-modules, TFAE: (i) M ⊂ E is a maximal essential extension. (ii) M ⊂ E is essential and E is injective. (iii) E is minimal injective over M : there does not exist any proper subextension M ⊂ E ′ ⊂ E with E ′ injective. Proof. We have already seen that (i) ⇐⇒ (ii). (ii) =⇒ (iii): Assume that E is injective and E ′ is an injective subextension of M ⊂ E . Since E ′ is injective, there exists N ⊂ E such that E ′ ⊕ N = E . Moreover, M ∩ N ⊂ E ′ ∩ N = 0, so M ∩ N = 0. Since M ⊂ E is essential, we must have N = 0, i.e., E ′ = E . (iii) =⇒ (ii): Suppose that M ⊂ E is minimal injective. The proof of Theorem 39 gives us a subextension E ′ of M ⊂ E such that E ′ is injective and M ⊂ E ′ is essential. Thus by minimality E = E ′ , i.e., M ⊂ E is essential. Theorem 42. (Uniqueness of Injective Envelopes) Let M be an R-module and let ι1 : M ⊂ E1 , ι2 : M ⊂ E2 be two injective envelopes of M . Then E1 and E2 are isomorphic as R-module extensions of M : i.e., there exists an R-module isomorphism Φ : E1 → E2 such that Φ ◦ ι1 = ι2 . Proof. Since ι1 : M → E1 is a monomorphism and E2 is injective, the map ι2 : M → E2 extends to a map Φ : E1 → E2 such that Φ ◦ ι1 = ι2 . Since the restriction of Φ to the essential submodule M is a monomorphism, so is Φ. The image Φ(E1 ) is an essential subextension of M ⊂ E2 , so by condition (iii) of Proposition 41 we must have E2 = Φ(E1 ). Thus Φ : E1 → E2 is an isomorphism. In view of Theorem 42, it is reasonable to speak of “the” injective envelope of M and denote it by M → E (M ). Reasonable, that is, but not ideal: it is not true that COMMUTATIVE ALGEBRA 45 any two injective envelopes are canonically isomorphic.15 Otherwise put, formation of the injective envelope is not functorial. For more on this in a more general category-theoretic context, see [AHRT02]. Exercise: Let M be a submodule of an injective module E . Show that E contains an isomorphic copy of the injective envelope E (M ). Exercise: If M ⊂ N is an essential extension of modules, then E (M ) = E (N ). 3.7. Flat modules. Suppose we have a short exact sequence 0 → M ′ → M → M ′′ → 0 of R-modules. If N is any R-module, we can tensor each element of the sequence with N , getting by functoriality maps 0 → M ′ ⊗ N → M ⊗ N → M ′′ ⊗ → 0. Unfortunately this new sequence need not be exact. It is easy to see that it is right exact: that is, the piece of the sequence M ′ ⊗ N → M ⊗ N → M ′′ ⊗ N → 0 remains exact. This follows because of the canonical “adjunction” ismorphism Hom(M ⊗ N, P ) = Hom(M, Hom(N, P )) and the left-exactness of the sequence Hom( , Y ) for all R-modules Y . However, tensoring an injection need not give an injection. Indeed, consider the exact sequence [2] 0 → Z → Z. If we tensor this with Z/2Z, we get a sequence [2] 0 → Z/2Z → Z/2Z, but now the map Z ⊗ Z/2Z → Z ⊗ Z/2Z takes n ⊗ i → (2n ⊗ i) = n ⊗ 2i = 0, so is not injective. Definition: A module M over a ring R is flat if the functor N → N ⊗R M is exact. This means, equivalently, that if M → M ′ then M ⊗ N → M ′ ⊗ N , or also that tensoring a short exact sequence with M gives a short exact sequence. It will probably seem unlikely at first, but in fact this is one of the most important and useful properties of an R-module. So, which R-modules are flat? Proposition 43. Let {Mi }i∈I be a family of R-modules. TFAE: (i) For all i, Mi is flat. ⊕ (ii) The direct sum M = i Mi is flat. 15The situation here is the same as for “the” splitting field of an algebraic field extension or “the” algebraic closure of a field. 46 PETE L. CLARK Exercise: Prove Proposition 43. Proposition 44. Let R be a domain. Then flat R-modules are torsionfree. Proof. We will prove the contrapositive. Suppose that 0 ̸= m ∈ R[tors], and let 0 ̸= r ∈ R be such that rm = 0 Since R is a domain, we have a short exact sequence [r ] 0 → R → R → R/rR → 0 and tensoring it with M gives [r ] 0 → M → M → M/rM → 0, but since rm = 0 the first map is not injective. Proposition 45. Projective R-modules are flat. Proof. A projective R-module is a module P such that there exists P ′ with P ⊕ P ′ ∼ = F a free module. Therefore, by Proposition 43, it is enough to show that free modules are flat. By abuse of notation, we will abbreviate the infinite direct sum of d copies of R as Rd . Since for any R-module M we have M ⊗R Rd = M d , it follows that tensoring a short exact sequence 0 → M ′ → M → M ′′ → 0 with F = Rd just yields 0 → (M ′ )d → (M )d → (M ′′ )d → 0. This is still exact. 3.8. Nakayama’s Lemma. 3.8.1. Nakayama’s Lemma. Proposition 46. Let M be a finitely generated R-module, I an ideal of R, and φ be an R-endomorphism of M such that φ(M ) ⊂ IM . Then φ satisfies an equation of the form φn + an−1 φn−1 + . . . + a1 φ + a0 = 0, with ai ∈ I . Proof. Let x1 , . . . , xn be a set of generators for M as an R-module. Since each ∑ φ(xi ) ∈ IM , we may write φ(xi ) = j aij xj , with aij ∈ I . Equivalently, for all i, n ∑ (δij φ − aij )xj = 0. j =1 By multiplying on the left by the adjoint of the matrix M = (δij φ − aij ), we get that det(δij φ − aij ) kills each xi , hence is the zero endomorphism of M . Expanding out the determinant gives the desired polynomial relation satisfied by φ. Exercise X.X: Some refer to Prop. 46 as the Cayley-Hamilton Theorem. Discuss. Theorem 47. (Nakayama’s Lemma) Let R be a ring, J an ideal of R, and M a finitely generated R-module such that JM = M . a) There exists x ∈ R with x ≡ 1 (mod J ) such that xM = 0. b) Suppose moreover that J is contained in every maximal ideal of R. Then M = 0. COMMUTATIVE ALGEBRA 47 Proof. Applying Proposition 46 to the identity endomorphism φ: gives a1 , . . . , an ∈ J such that for x := 1 + a1 + . . . + an , xM = 0 and x ≡ 1 (mod J ), proving part a). If moreover J lies in every maximal ideal m of R, then x ≡ 1 (mod )m for all maximal ideals m, hence x lies in no maximal ideal of R. Therefore x is a unit and xM = 0 implies M = 0. Corollary 48. Let R be a ring, J an ideal of R which is contained in every maximal ideal of R, M a finitely generated R-module and N a submodule of M such that JM + N = M . Then M = N . Proof. We have J (M/N ) = (JM + N )/N = M/N . Applying Nakayama’s Lemma to the finitely generated module M/N , we conclude M/N = 0, i.e., N = M . Corollary 49. Let R be a ring, J an ideal of R which is contained in every maximal ideal of R, and M a finitely generated R-module. Let x1 , . . . , xn ∈ M be such that their images in M/JM span M/JM as an R/J -module. Then the xi ’s span M . Proof. Let N = ⟨x1 , . . . , xn ⟩R , and apply Corollary 48. Corollary 50. Let R be a ring and J an ideal which is contained in every maximal ideal of R. Let M and N be R-modules, with N finitely generated, and let u : M → N be an R-module map. Suppose that the map uJ : M/JM → N/JN is surjective. Then u is surjective. Proof. Apply Nakayama’s Lemma to J and N/M . Recall that an element x in a ring R such that x2 = x is called idempotent. Similarly, an ideal I of R such that I 2 = I is called idempotent. Exercise: Let R be a ring and I an ideal of R. a) Suppose I = (e) for an idempotent element e. Show that I is idempotent. b) Give an example of a nonidempotent x such that (x) is idempotent. c) Is every idempotent ideal generated by some idempotent element? The last part of the preceding exercise is rather difficult. It turns out that the answer is negative in general: we will see later that counterexamples exist in any infinite Boolean ring. However under a relatively mild additional hypothesis the answer is affirmative. Corollary 51. Let R be any ring and I a finitely generated idempotent ideal of R. Then there exists an idempotent e ∈ R such that I = (e). In particular, in a Noetherian ring every idempotent ideal is generated by a single idempotent element. Exercise: Prove Corollary 51. (Hint: apply Theorem 47!) 3.8.2. Hopfian modules. A group G is Hopfian if every surjective group homomorphism f : G → G is an isomorphism – equivalently, G is not isomorphic to any of its proper quotients. This concept has some currency in combinatorial and geometric group theory. Some basic examples: any finite group is certainly Hopfian. A free group is Hopfian iff it is finitely generated, and more generally a finitely generated residually finite 48 PETE L. CLARK ∏∞ group is Hopfian. An obvious example of a non-Hopfian group is i=1 G for any nontrivial group G. A more interesting example is the Baumslag-Solitar group B (2, 3) = ⟨x, y | yx2 y −1 = x3 ⟩. More generally, let C be a concrete category: that is, Ob C is a class of sets and for all X, Y ∈ Ob C , HomC (X, Y ) ⊂ HomSet (X, Y ), i.e., the morphisms between X and Y are certain functions from X to Y . We may define an object X in C to be Hopfian if every surjective endomorphism of X is an isomorphism. Exercise X.X: a) Give an example of a concrete category C and a finite object X (i.e., the underlying set of X is finite) which is not Hopfian. b) Suppose that C satisfies the property that any bijective morphism is an isomorphism. Show that any finite object is a Hopfian object. c) In the category of Sets, the Hopfian objects are precisely the finite sets. Remark: Our discussion of “Hopfian objects” in categories more general than RMod is not particularly serious or well thought out. So far as I know there is not a completely agreed upon definition of a Hopfian object, but Martin Brandenburg has suggested (instead) the following: X ∈ C is Hopfian if every extremal epimorphism X → X is an isomorphism. Theorem 52. Let R be a ring and M a finitely generated R-module. Then M is a Hopfian object in the category of R-modules. Proof. ([M, p. 9]) Let f : M → M be a surjective R-map. We show f is injective. There is a unique R[t]-module structure on M extending the given R-module structure and such that for all m ∈M, tm = f (m). Let I = tR[t]. By hypothesis IM = M , so by Nakayama’s Lemma there exists P (t) ∈ R[t] such that (1 + P (t)t)M = 0. Let y ∈ ker f . Then 0 = (1 + P (t)t)y = y + P (t)f (y ) = y + P (t)0 = y. So f is injective. Exercise: Show that (Q, +) is a Hopfian Z-module which is not finitely generated. Exercise: Do there exist Hopfian Z-modules of all cardinalities? (An affirmative answer was claim in [Bau62], but it was announced in [Bau63] that the construction is not valid. So far as I know the problem remains open lo these many years later.) 3.8.3. A variant. The results of this section are taken from [DM71, §I.1]. Proposition 53. (Generalized Nakayama’s Lemma) Let R be a ring, J an ideal of R and M a finitely generated R-module. TFAE: (i) J + ann M = R. (ii) JM = M . Proof. (i) =⇒ (ii): If J + ann M = R, we may write 1 = x + y with x ∈ J, y ∈ ann M , so that for all m ∈ M , m = 1m = xm + ym = xm. Thus JM = M . (ii) =⇒ (i): Conversely, suppose M = ⟨m1 , . . . , mn ⟩. For 1 ≤ i ≤ n, put COMMUTATIVE ALGEBRA 49 Mi = ⟨mi , . . . , mn ⟩ and Mn+1 = 0. We claim that for all 1 ≤ i ≤ n + 1 there exists ai ∈ J with (1 − ai )M ⊂ Mi , and we will prove this by induction on n. We may take a1 = 0. Having chosen a1 , . . . , ai , we have (1 − ai )M = (1 − ai )JM = J (1 − ai )M ⊂ Mi , so there exist aij ∈ J such that (1 − ai )mj = n ∑ aij mj , j =i or (1 − ai − aii )mi ∈ Mi+1 . Thus ( ) 1 − (2ai + aii − a2 − ai ai i) M = (1 − ai )(1 − ai − aii )M ⊂ (1 − ai − aii )Mi ⊂ Mi+1 , i and we may take ai+1 = 2ai + aii − a2 − ai ai i. i So there is an ∈ J such that 1 − an ∈ ann M , and thus 1 ∈ J + ann M . Exercise: Deduce part b) of Nakayama’s Lemma from Proposition 53. Corollary 54. Let M be a finitely generated R-module such that mM = M for all maximal ideals of R. Then M = 0. Exercise: Prove Corollary 54. For an R-module M , we define its trace ideal to be the ideal T (M ) of R generated by all the images f (M ) of R-module maps f ∈ R∨ = HomR (M, R). Theorem 55. Let P be a finitely generated projective R-module. Then R splits as a direct product of rings: R = T (P ) ⊕ ann P. Proof. Step 1: We show that T (P ) and ann P are comaximal ideals of R, i.e., T (P ) + ann P = R. By the Dual Basis Lemma (Proposition 20) and the following exercise, there exist x1 , . . . , xn ∈ P∑ and f1 , . . . , fn ∈ P ∨ = HomR (P, R) forming n a dual basis: for all x ∈ P , x = i=1 fi (x)xi . By its very definition we have fi (x) ∈ T (P ) for all i and x, hence T (P )P = P . By the Generalized Nakayama’s Lemma (Lemma 53) we have T (P ) + ann P = R. For any a ∈ ann P , f ∈ P ∨ and x ∈ P we have af (x) = f (ax) = f (0) = 0: thus T (P ) ∩ ann P = 0. By comaximality, T (P ) ∩ ann P = 0 and the sum is direct. Corollary 56. A nonzero finitely generated projective module over a connected ring R (i.e., without idempotents other than 0 and 1) is faithful. Exercise: Prove Corollary 56. 50 PETE L. CLARK 3.8.4. Applications to modules over local rings. Lemma 57. Let R be a ring and J an ideal which is contained in every maximal ideal of R, and let M be a finitely presented R-module. Suppose that: (i) M/JM is a free R/J -module, and (ii) The canonical map J ⊗R M → JM is injective. Then M is a free R-module. Proof. We may choose a family {xi }i∈I of elements of M such that the images in M/JM give a R/J -basis. (Since M is finitely generated over R, M/JM is finitely generated over R/J , so the index set I is necessarily finite.) Consider the finitely ⊕ generated free R-module L = i∈I R, with canonical basis {ei }. Let u : L → M be the unique R-linear mapping each ei to xi , and let K = ker(u). Since M is finitely presented, by Proposition 14 K is finitely generated. We have a commutative diagram with exact rows: J ⊗K →J ⊗L→J ⊗M →0 0 → K → L → M → 0, where each vertical map – a : J ⊗ K → K , b : J ⊗ L → L, c : J ⊗ M → M – is the natural multiplication map. Our hypothesis is that the the map J ⊗R M → JM is injective, so by the Snake Lemma we get an exact sequence u 0 → coker(a) → coker(b) → coker(c). Now observe that coker(b) = (R/J ) ⊗R L and coker(c) = (R/J ) ⊗R M , and by definition the mapping u : L → M gives, upon passage to the quotient modulo J , a mapping from one R/J -module basis to another. So u is an isomorphism and thus coker(a) = 0, i.e., K/JK = 0. By Nakayama’s Lemma we conclude K = 0, i.e., u gives an isomorphism from the free module L to M , so M is free. We can now prove the following result, which is one that we will build upon in our future studies of modules over commutative rings. Theorem 58. Let R be a ring with a unique maximal ideal m – i.e., a local ring. For a finitely presented R-module M , TFAE: (i) M is free. (ii) M is projective. (iii) M is flat. (iv) The natural map m ⊗R M → mM is an injection. Proof. Each of the implications (i) =⇒ (ii) =⇒ (iii) =⇒ (iv) is immediate. Assume (iv). Then, since m is maximal, R/m is a field, so every R/m-module is free. Therefore Lemma 57 applies to complete the proof. 3.9. Ordinal Filtrations and Applications. 3.9.1. The Transfinite D´vissage Lemma. e Let M be an R-module. By an ordinal filtration on M we mean an ordinal number α and for each i ≤ α a submodule Mi of M satisfying all of the following: (OF1) M0 = 0, Mα = M . (OF2) For all i, j ∈ α + 1, i ≤ j =⇒ Mi ∪ Mj . ⊂ (OF3) For all limit ordinals i ≤ α, Mi = j<i Mj . COMMUTATIVE ALGEBRA 51 So for instance, taking α = ω = {1, 2, 3, . . .} the first infinite ordinal, we recover the usual notion of an exhaustive filtration by submodules Mn , with the additional ∪ convention that Mω = n∈ω Mn . For i < α, we call Mi+1 /Mi the ith successive quotient. If for a class C of R-modules each successive quotient lies in C , we say the filtration is of class C . Define the associated graded module Gr(M ) = ⊕ i<α Mi+1 /Mi . Lemma 59. (Transfinite D´vissage Lemma) Let M be an R-module and {Mi }i≤α e an ordinal filtration of M . a) Suppose we make the following hypothesis: (DS) For all i < α the submodule Mi is a direct summand of Mi+1 . Then ⊕ M ∼ Gr(M ) = Mi+1 /Mi . = i<α b) Hypothesis (DS) holds if each successive quotient Mi+1 /Mi is projective. c) Hypothesis (DS) holds if each Mi is injective. Exercise: Prove Lemma 59. (Hint: transfinite induction.) Corollary 60. For an R-module M , TFAE: (i) M is free. (ii) M admits an ordinal filtration with successive quotients isomorphic to R. (iii) M admits an ordinal filtration with free successive quotients. ⊕ Proof. (i) =⇒ (ii): If M is free, then M ∼ = i∈I R. By the Well-Ordering ⊕ Principle16, I is in bijection with an ordinal α, so we may write M ∼ i<α R, and = ⊕ put Mi = j<i R. (ii) =⇒ (iii) is immediate. (iii) =⇒ (i) follows from Lemma 59 since free modules are projective. 3.9.2. Hereditary and semihereditary rings. An R-module is hereditary if every submodule is projective. (In particular a hereditary module is projective, and thus the property of being projective is “inherited” by its subomdules.) We say that a ring R is hereditary if R is a hereditary R-module, or equivalently every ideal of R is projective as an R-module. Exercise: a) Show that every submodule of a hereditary module is hereditary. b) Show that the zero module is hereditary. c) Show that there are nonzero rings R for which the only hereditary R-module is the zero module. Example: A PID is a hereditary ring. Indeed, any nonzero ideal of a PID R is isomorphic as an R-module to R. 16Recall this is a set-theoretic axiom which is equivalent to the Axiom of Choice and also to Zorn’s Lemma and that our running convention in these notes to freely use these axioms when necessary. 52 PETE L. CLARK Vista: We will not study Dedekind domains until much later on, but (especially, number-theoretically inclined) readers may nevertheless be familiar with them. It turns out that any Dedekind domain is a hereditary ring and conversely a hereditary domain is necessarily a Dedekind domain. ⊕ Theorem 61. Let {Mi }i∈I a family of hereditary R-modules, put M = i∈I Mi , and let πi : M → Mi be projection onto the ith factor. Then, for any submodule N ⊕ of M , N ∼ i∈I πi (N ). = Proof. By the Well-Ordering Principle there exists a bijection from I to some or⊕ dinal α, and without loss of generality we may assume M = i∈α Mi . For j ∈ α+ , ⊕ put Pj = i<j + Mi , so that {Mj } is an ordinal-indexed chain of R-submodules of M with final element Pα = M . For each j ∈ α+ , put Nj = N ∩ Pj , so {Nj } is an ordinal filtration on N with Nα = N . Moreover, for all i ∈ α we have Ni = Ni+1 ∩ Pi and thus ∼ Ni+1 /Ni = Ni /(Ni+1 ∩ Pi ) = (Ni+1 + Pi )/Pi . Thus Ni+1 /Ni is isomorphic to a submodule of Pi+1 /Pi ∼ Mi . Since each Mi = is hereditary, each successive quotient Ni+1 /Ni is projective, and the Transfinite D´vissage Lemma (Lemma 59) applies to show that e ⊕ N ∼ Gr N = Nj +1 /Nj = = ⊕ j<α+ (N ∩ j<α+ ⊕ i<j ++ Mj )/(N ∩ ⊕ i<j + Mj ) ∼ = ⊕ πj (N ). j<α+ Corollary 62. Let {Mi }i∈I be a family of R-modules. Then M = hereditary iff Mi is hereditary for all i. ⊕ i ∈I Mi is Proof. Suppose first that each Mi is hereditary, let N be a submodule of M , and ⊕ put Ni = πi (N ). By Theorem 61, N ∼ i∈I Ni . Each Ni is a submodule of the = hereditary module M hence is projective. Thus N is a direct sum of projective modules, hence projective. Conversely, if M is hereditary, then so are all of its submodules, but the canonical map ιi : Mi → M realizes Mi as a submodule of M . Corollary 63. For a ring R, TFAE: (i) R is hereditary. (ii) Every free R-module is hereditary. (iii) Every projective R-module is hereditary. Proof. (i) =⇒ (ii) is immediate from Corollary 62. (ii) =⇒ (iii): Suppose that every free R-module is hereditary. Then if P is a projective R-module, P is a submodule of a free module, hence a submodule of a hereditary module, hence itself heeditary. (iii) =⇒ (i): R is a projective R-module. COMMUTATIVE ALGEBRA 53 ⊕ Theorem 64. Let R be a a PID and F = i∈I R a free R-module. Then any submodule M of F is again free, of rank less than or equal to the rank of F . ⊕ Proof. Let N be a submodule of F . By Theorem 61, N ∼ i∈I σi (N ), where each = σi (N ) is an R-submodule of R, i.e., an ideal of R. Since R is a PID, either σi (N ) is the zero module or is isomorphic as an R-module to R. Done! Corollary 65. A projective module over a PID is free. Proof. A projective module is a direct summand of a free module, and in particular is a submodule of a free module. Now Theorem 64. We expect that the following result is familiar to the reader as a special case of the classification of (all) finitely generated modules over a PID, but while we are here we may as well give a commutative algebraic proof. Proposition 66. Let R be a PID. A finitely generated torsionfree R-module is free. Proof. Let M be finitely generated and torsionfree. Certainly we may assume that M is nonzero. Let X be a finite generating set for M with 0 ∈ X . Let S ⊂ X be / a maximal R-linearly independent subset. Since M is torsionfree, S is not empty. Let N = ⟨S ⟩R be the R-module spanned by S . Clearly N is free with basis S , and we will be done if we can show N = M . There is an annoying technicality here: we must check that S is finite. In fact, let n = #X , and let S be any finite R-linearly independent subset and let s = #S . We claim that s ≤ n. To see this, we tensor with the fraction field K of R, getting ιK K s ∼ S ⊗R K → M ⊗R K ∼ K m with m ≤ n. We may conclude that s ≤ m ≤ n = = provided we know that ιK is injective. And this holds because K is a flat R-module, which we will prove later on as a special case of the flatness of localization maps. Until then, the reader must take this part of the proof on faith. Say S = {x1 , . . . , xk }. Of course we are done already if S = X , so assume that X \ S is nonempty. For each y ∈ X \ S there exist ry , r1 , . . . , rk ∈ R, not all zero, such that ry y = r1 x1 + . . . + rk xk . Then ry ̸= 0, since otherwise by the linear independence of S all the ri would be zero. In other words, we have ∏ shown that for each y ∈ X \ S there exists ry ∈ R• such that ry y ∈ N . Put r := y∈X \S ry . Then rX ⊂ N and thus rM ⊂ N . Now consider the R-module homomorphism L : M → M given by multiplication by r: x → rx. We have just established that L(M ) ⊂ N , so we may regard L as a homomorphism M → N . Moreover, since M is torsionfree, L is injective, and therefore L realizes M as a submodule of the free R-module N . By Theorem 64 we conclude that M is free. Exercise X.X: Let R be a ring with the following property: every submodule of a finitely generated free R-module is free. Show that R is a principal ring (i.e., every ideal of R is principal). An R-module M is semihereditary if every finitely generated submodule is projective. Thus a Noetherian semihereditary module is hereditary. A ring R is semihereditary if R is a semihereditary R-module, or equivalently every finitely generated ideal of R is projective as an R-module. Example: A domain R is semihereditary if every finitely generated ideal is principal. Such domains are called B´zout domains and will be further studied later e on. 54 PETE L. CLARK ⊕ Theorem 67. Let {Mi }i∈I a family of semihereditary R-modules, put M = i∈I Mi , and let πi : M → Mi be projection onto the ith factor. Then, for any finitely gen⊕ erated submodule N of M , N ∼ i∈I πi (N ). = Proof. The proof of Theorem 61 goes through verbatim. Theorem 68. Let R be a B´zout domain – i.e., a domain in which every finitely e generated ideal is principal – and F a free R-module. Then any finitely generated submodule N of F is free, of rank less than or equal to the rank of F . Proof. The proof of Theorem 65 adapts immediately, using Theorem 67 in place of Theorem 61. Theorem 69. Let R be a B´zout domain – i.e., a domain in which every finitely e generated ideal is principal. Then every finitely generated torsionfree R-module is principal. Proof. The argument is the same as that of Proposition 66 (a special case), using Theorem 68 in place of Theorem 64. Theorem 70. (F. Albrecht) Let R be a semihereditary ring, F a free R-module, and P a finitely generated submodule of F . a) P is isomorphic to a finite direct sum of finitely generated ideals of R. b) In particular, P is a finitely generated projective module. c) If R is a domain with fraction field K and F is free of finite rank n, then the rank of P – i.e., dimK P ⊗R K – is at most n. Exercise: Use Theorem 67 to prove Theorem 70. 3.9.3. Big modules. Lemma 71. (Kaplansky) Let R be a ring, and let F be an R-module which is a ⊕ direct sum of countably generated submodules: say F = λ∈Λ Eλ . Then every direct summand of F is again a direct sum of countably generated submodules. Proof. We claim that there is an ordinal filtration {Fi }i≤α on F satisfying all of the following properties. (i) For all i < α, Fi+1 /Fi is countably generated. (ii) If Mi = Fi ∩ M, Ni = Fi ∩ N , then Fi = Mi ⊕ Ni . ⊕ (iii) For each i there is a subset Λi of Λ such that Fi = λ∈Λi Λi . sufficiency of claim: If so, then {Mi }i≤α is an ordinal filtration on M . Moreover, for all i, since Mi ⊂ Mi+1 are both direct summands of F , Mi is a direct summand of Mi+1 . Therefore the Transfinite D´vissage Lemma (Lemma 59) applies to the filtration e on M to give ⊕ Mi+1 /Mi . M ∼ Gr(M ) = = i<α Moreover, for all i < α we have Fi+1 /Fi = (Mi+1 ⊕ Ni+1 )/(Mi ⊕ Ni ) ∼ Mi+1 /Mi ⊕ Ni+1 /Ni , = which shows that each successive quotient Mi+1 /Mi is countably generated. Therefore M is a direct sum of countably generated submodules. proof of claim: We will construct the filtration by transfinite induction. The base case and the limit ordinal induction step are forced upon us by the definition of ordinal filtration: we must have F0 = {0}, and for any limit ordinal β ≤ α, ∪ assuming we have defined Fi for all i < β we must have Fβ = i<β Fi . COMMUTATIVE ALGEBRA 55 So consider the case of a successor ordinal β = β ′ + 1. Let Q1 be any Eλ which is not contained in Fβ ′ . (Otherwise we have Fβ ′ = F and we may just define Fi = F for all β ≤ i ≤ α.) Let x11 , x12 , . . . be a sequence of generators of Q1 , and decompose x11 into its M - and N -components. Let Q2 be the direct sum of the finitely many Eλ which are necessary to write both of these components, and let x21 , x22 , . . . be a sequence of generators for Q2 . Similarly decompose x12 into M and N components, and let Q3 be the direct sum of the finitely many Eλ needed to write out these components, and let x31 , x32 , . . . be a sequence of generators of Q3 . We continue to carry out this procedure for all xij , proceeding according to a diagonal enumeration of Z+ × Z+ : i.e., x11 , x12 , x21 , x13 , x22 , x31 , . . .. Put Fβ = ⟨Fβ ′ , {xij }i,j ∈Z+ ⟩R . This works! For a cardinal number κ, we say that a module is κ-generated if it admits a generating set of cardinality at most κ. Exercise (Warfield): Let κ be any infinite cardinal. Formulate and prove a version of Lemma 71 in which each instance of “countably generated” is replaced by κ-generated. Theorem 72. (Kaplansky) For a ring R, let Pc be the class of countably generated projective R-modules. For an R-module M , TFAE: (i) M admits an ordinal filtration of class Pc . (ii) M is a direct sum of countably generated projective submodules. (iii) M is projective. Proof. (i) ⇐⇒ (ii) follows immediately from Lemma 59. (ii) =⇒ (iii): any direct sum of projective modules is projective. (iii) =⇒ (ii): If M is projective, let F be a free R-module with F = M ⊕ N . Certainly F is a direct sum of countably generated submodules (indeed, of singly generated submodules!), so by Lemma 71 M is a direct sum of a family of countably generated submodules, each of which must be projective. While pondering the significance of this result, one naturally inquires: Question 2. Is there a ring R and an R-module M which is not a direct sum of countably generated submodules? Theorem 73. (Cohen-Kaplansky [CK51], Griffith) For a ring R, TFAE: (i) Every R-module is a direct sum of cyclic (i.e., singly generated) R-modules. (ii) Every R-module is a direct sum of finitely generated submodules. (iii) R is an Artinian principal ideal ring. Building on these results as well as work of Faith and Walker [FW], R.B. Warfield Jr. proved the following striking results. Theorem 74. (Warfield [Wa]) Let R be a Noetherian ring which is not a principal Artinian ring. Then for any cardinal κ, there exists a module M with the following properties: (i) M is not κ-generated, and (ii) Any decomposition of M into the direct sum of nonzero submodules has only finitely many direct summands. 56 PETE L. CLARK The hypotheses of Theorem 74 apply for instance to the ring Z of integers and yields, in particular, for any infinite cardinal κ a commutative group M which is not a direct sum of κ-generated submodules. Theorem 75. (Warfield [Wa]) For a ring R, TFAE: (i) Every R-module is a direct sum of cyclic submodules. (ii) There exists a cardinal number κ such that every R-module is a direct sum of κ-generated submodules. (iii) R is a principal Artinian ring. It is natural to wonder whether Theorem 74 can be strengthened in the following way: an R-module M is indecomposable if it cannot be expressed as a direct sum of two nonzero submodules. Question 3. For which rings R do there exist indecomposable R-modules of all infinite cardinalities? However, Question 3 has turned out to be bound up with sophisticated set-theoretic considerations. Namely, in a 1959 paper [Fu59], L. Fuchs claimed that there exist indecomposable commutative groups of all infinite cardinalties, thus giving an affirmative answer to Question 3 for the ring R = Z. However, it was later observed (by A.L.S. Corner) that Fuchs’ argument is valid only for cardinals κ less than the first inaccessible cardinal. Exactly what an inaccessible cardinal is we do not wish to say, but we mention that the nonexistence of inaccessible cardinals is equiconsistent with the standard ZFC axioms of set theory (in other words, if the ZFC axioms are themselves consistent, then ZFC plus the additional axiom that there are no inaccessible cardinals remains consistent) but that nevertheless set theorists have reasons to believe in them. See also [Fu74] in which these issues are addressed and he proves that there is an indecomposable commutative group of any infinite nonmeasurable cardinality (note: accessible implies nonmeasurable). Question 4. Is there a ring R and a projective R-module M which is not a direct sum of finitely generated submodules? Again the answer is yes. A very elegant example was given by Kaplansky (unpublished, apparently).17 Namely that R be the ring of all real-valued continuous functions on the unit interval [0, 1], and let I be the ideal of functions f : [0, 1] → R which vanish near zero: i.e., for which there exists ϵ = ϵ(f ) > 0 such that f |[0,ϵ(f )] = 0. Exercise X.X: Show the ideal I defined above gives a projective R-module which is not the direct sum of finitely generated submodules. (Suggestions: (i) to show that I is projective, use the Dual Basis Lemma. (ii) A slick proof of the fact that I is not a direct sum of finitely generated submodules can be given by Swan’s Theorem using the contractibility of the unit interval.) Lemma 76. Let M be a projective module over the local ring R, and let x ∈ M . There is a direct summand M ′ of M such that M ′ contains F and M ′ is free. Proof. Let F be a free module with F = M ⊕ N . Choose a basis B = {ui }i∈I of F with respect to which the element x of M has the minimal possible number of nonzero coordinates. Write x = r1 u1 + . . . + rn un , ri ∈ R• . 17Warm thanks to Gjergji Zaimi for bringing this important example to my attention. COMMUTATIVE ALGEBRA 57 ∑ ∑n−1 Then for all 1 ≤ i ≤ n, ri ∈ / j ̸=i Rrj . Indeed, if say rn = i=1 si ri , then ∑n−1 x = i=1 ri (ui + si un ), contradicting the minimality of the chosen basis. Now write ui = yi + zi with yi ∈ M, zi ∈ N , so ∑ ∑ (2) x= ri u i = ri yi . i i We may write (3) yi = n ∑ cij uj + ti , j =1 with ti a linear combination of elements of B \ {u1 , . . . , un }. Substituting (3) into (2) and projecting onto M gives the relations ri = n ∑ cji rj , j =1 or equivalently, for all i, (1 − cii )ri = ∑ cji rj . j ̸=i If for any i and j , then one of the coefficients of rj in the above equation is a unit of R, then dividing through by it expresses rj as an R-linear combination of the other ri ’s, which as above is impossible. Therefore, since R is local, each coefficient must lie in the maximal ideal of R: ∀i, 1 − cii ∈ m, ∀i ̸= j, cij ∈ m. It follows that the determinant of the matrix C = (cij ) is congruent to 1 modulo m, hence invertible: if x ∈ m and 1 + x is not invertible, then 1 + x = y for y ∈ m, so 1 = y − x ∈ m, contradiction. Therefore replacing u1 , . . . , un in B with y1 , . . . , yn still yields a basis of F . It follows that M ′ = ⟨y1 , . . . , yn ⟩R is a direct summand of F hence also of M which is a free module containing x. Theorem 77. (Kaplansky) Let (R, m) be a local ring, and let P be any projective R-module. Then P is free. Proof. Step 1: Since by Theorem 72 P is a direct sum of countably generated projective submodules, we may as well assume that P itself is countably generated. Step 2: Suppose M = ⟨{xn }∞ ⟩R is a countably generated projective module over n=1 the local ring R. By Lemma 76, M = F1 ⊕ M1 with F1 free containing x1 . Note that M1 is again projective and is generated by the images {x′ }∞ of the elements n n=2 xn under the natural projection map M → M1 . So reasoning as above, we may write M2 = F2 ⊕ M2 with F2 free containing x′ . Continuing in this manner, we get 2 M= ∞ ⊕ Fn , n=1 so M is free. Exercise X.X: Give an example of a (necessarily infinitely generated) module over a local PID which is flat but not free. 3.10. Tor and Ext. 58 PETE L. CLARK 3.10.1. Co/chain complexes. Let R be a ring. A chain complex C• of R-modules is a family {Cn }n∈Z of R-modules together with for all n ∈ Z, an R-module map dn : Cn → Cn−1 such that for all n, dn−1 ◦ dn = 0. (It is often the case that Cn = 0 for all n < 0, but this is not a required part of the definition.) An example of a chain complex of R-modules is any long exact sequence. However, from the perspective of homology theory this is a trivial example in the following precise sense: for any chain complex we may define its homology modules: for all n ∈ Z, we put Hn (C ) = Ker(dn )/ Im(dn+1 ). Example: Let X be any topological space. For any ring R, we have the singular chain complex S (X )• : S (X )n = 0 for n < 0, and for n ≥ 0, S (X )n is the free R-module with basis the set of all continuous maps ∆n → X , where ∆n is the standard n-dimensional simplex. A certain carefully defined alternating sum of restrictions to faces of ∆n gives rise to a boundary map dn : S (X )n → S (X )n−1 , and the indeed the homology groups of this complex are nothing else than the singular homology groups Hn (X, R) with coefficients in R. If C• and D• are two chain complexes of R-modules, a homomorphism η : C• → D• is given by maps ηn : Cn → Dn for all n rendering the following infinite ladder commutative: INSERT ME!. In this way one has evident notions of a monomorphism and epimorphisms of chain complexes. In fact the chain complexes of R-modules form an abelian category and thus these notions have a general categorical meaning, but it turns out they are equivalent to the much more concrete naive conditions: η is a monomorphism iff each ηn is injective and is an epiomorphism iff each ηn is surjective. In particular it makes sense to consider a short exact sequence of chain complexes: 0 −→ A• −→ B• −→ C• . Here is the first basic theorem of homological algebra. Theorem 78. Let f g 0 −→ A• −→ B• −→ C• −→ 0 be a short exact sequence of chain complexes of R-modules. Then for all n ∈ Z there is a natural connecting homomorphism ∂ : Hn (C ) → Hn−1 (A) such that g ∂ f g ∂ f . . . −→ Hn+1 (C ) −→ Hn (A) → Hn (B ) −→ Hn (C ) −→ Hn−1 (A) −→ . . . is exact. Proof. No way. See [W, Thm. 1.3.1]. Moreover, the homology modules Hn are functors: if f : C• → D• is a morphism of chain complexes, there are induced maps on the homology groups Hn (f ) : Hn (C ) → Hn (D). Example: Let f : X → Y be a continuous map of topological spaces. Then for any basic n-chain ∆n → X in S (X )n , composition with f gives a basic n-chain ∆n → Y COMMUTATIVE ALGEBRA 59 in S (Y )n and thus a homomorphism of chain complexes S (f ) : S (X )• → S (Y )• . There are induced maps on homology, namely the usual maps Hn (f ) : Hn (X, R) → Hn (Y, R). There is an entirely parallel story for cochain complexes of R-modules, which are exactly the same as chain complexes but with a different indexing convention: a cochain complex C • consists of for each n ∈ Z+ an R-module C n and a “coboundary map” dn : C n → C n+1 . To any cochain complex we get cohomology modules: for all n ∈ Z, put H n (C ) = Ker(dn )/ Im(dn−1 ). The rest of the discussion proceeds in parallel to that of chain complexes (including the realization of singular cohomology as a special case of this construction). 3.10.2. Chain homotopies. Let C• , D• be two chain complexes, and let f, g : C• → D• be two homomorphisms between them. We say that f and g are chain homotopic if there exist for all n ∈ Z+ R-module maps sn : Cn → Dn+1 such that fn − gn = dn+1 sn + sn−1 dn . The sequence {sn } is called a chain homotopy from f to g . Exercise X.X: Show that chain homotopy is an equivalence relation on morphisms from C• to D• . What on earth is going on here? Again topology is a good motivating example: we say that two maps f, g : X → Y are homotopic if there exists a continuous map F : X × [0, 1] → Y such that for all x ∈ X , F (x, 0) = f (x) and F (x, 1) = g (x). This is an equivalence relation and is generally denoted by f ∼ g . We then define two topological spaces to be homotopy equivalent if there exist maps φ : X → Y and ψ : Y → X such that ψ ◦ φ ∼ 1X , φ ◦ ψ ∼ 1Y . (We say that φ : X → Y is a homotopy equivalence if there exists a map ψ as above.) E.g. a space is contractible if it is homotopy equivalent to a single point. One of the basic tenets of algebraic topology is that it aspires to study topological spaces only up to homotopy equivalence. That is, all of the fundamental invariants of spaces should be the same on homotopy equivalent spaces and homomorphisms between these invariants induced by homotopic maps should be identical. Especially, if f : X → Y is a homotopy equivalence, the induced maps Hn (f ) : Hn (X ) → Hn (Y ) should be isomorphisms. In fact, if f, g : X → Y are homotopic, the induced morphisms S (f ), S (g ) : S (X )• → S (Y )• are chain homotopic. So the following result ensures that the induced maps on homology are equal. Proposition 79. If f, g : C• → D• are chain homotopic, then for all n ∈ Z, Hn (f ) = Hn (g ). Proof. Replacing f and g by f − g and 0, it is enough to assume that there exists a chain homotopy s from f to the zero map – i.e., for all n fn = dn+1 sn + sn−1 dn 60 PETE L. CLARK – and show that f induces the zero map on homology. So take x ∈ Hn (C ). Then x is represented by an element of Cn lying in the kernel of dn , so fn (x) = dn+1 sn x + sn−1 dn x = dn+1 sn x + 0 = dn+1 sn x. Thus fn (x) lies in the image of dn+1 Dn+1 → Dn so represents 0 ∈ Hn (D). 3.10.3. Resolutions. Let M be an R-module. A left resolution of M is an infinite sequence {Ai }∞ i=0 of R-modules, for all n ∈ N an R-module map An+1 → An and an R-module map A0 → M such that the sequence . . . −→ An+1 → An −→ . . . −→ A1 −→ A0 −→ M −→ 0 is exact. By abuse of notation, we often speak of “the resolution A• . Dually, a right resolution of M is an infinite sequence {B i }∞ of R-modules, for all n ∈ N i=0 an R-module map B n → B n+1 and an R-module map M → B 0 such that the sequence 0 −→ M −→ B 0 −→ B 1 −→ . . . −→ B n −→ B n1 . . . is exact. We speak of “the resolution B • ”. A projective resolution of M is a left resolution A• such that each An is projective. A injective resolution of M is a right resolution B • such that each B n is injective. (Exactly why we are not interested in left injective resolutions and right projective resolutions will shortly become clear.) Theorem 80. (Existence of resolutions) Let M be an R-module. a) Since every R-module is the quotient of a projective (indeed, of a free) module, M admits a projective resolution. b) Since every R-module can be embedded in an injective module, M admits an injective resolution. Proof. a) Choose a projective module P0 , a surjection ϵ0 : P0 → M , and put M0 = ker(ϵ0 ). Inductively, given Mn−1 , we choose a projective module Pn , a surjection ϵn : Pn → Mn−1 , and put Mn = ker(ϵ0 ). As our map dn : Pn → Pn−1 we take the composite ϵ n Pn −→ Mn−1 ker(ϵn−1 ) −→ Pn−1 . We claim that the resulting sequence . . . −→ Pn+1 → Pn −→ . . . −→ P1 −→ P0 −→ M −→ 0 is exact. It is certainly exact at M . If x ∈ P0 and ϵ0 (x) = 0, then x0 ∈ M0 . Lifting x0 via the surjection ϵ1 to x1 ∈ P1 , we find d1 (x1 ) = ϵ1 (x1 ) = x0 , so ker(ϵ0 ) ⊂ Im(d1 ). Conversely, since d1 factors through ker(ϵ0 ), it is clear that Im(d1 ) ⊂ ker(ϵ0 ). Exactly the same argument verifies exactness at Pn for each n > 0, so P• is a projective resolution of M . b) We leave the proof of this part to the reader as an exercise, with the following comforting remark: the notion of an injective module is obtained from the notion of a projective module by “reversing all the arrows”, which is the same relationship that a left resolution bears to a right resolution. Therefore it should be possible to prove part b) simply by holding up the proof of part a) to a mirror. (And it is.) COMMUTATIVE ALGEBRA 61 Theorem 81. (Comparison theorem for resolutions) a) Let P• be a projective resolution of the R-module M . Let N be another Rmodule and f ′ : M → N be an R-module map. Then for every left resolution A• of N there exists a homomorphism η from the chain complex P• → M → 0 to the chain complex A• → N → 0. Moreover η is unique up to chain homotopy. b) Let E • be an injective resolution of the R-module N . Let M be another R-module and f ′ : M → N be an R-module map. Then for every right resolution A• of M there exists a homomorphism η from the chain complex 0 → M → A• to the chain complex 0 → N → E • . Moreover η is unique up to chain homotopy. Proof. No way. See [W, Thms. 2.2.6 and 2.3.7]. Exercise: Let F be a covariant additive functor on the category of R-modules. Let C• and D• be two chain complexes of R-modueles and f, g : C• → D• be two homomorphisms between them. a) Show that F C• and F D• are chain complexes and there are induced chain homomorphisms F f, F g : F C• → F D• . b) Show that if f and g are chain homotopic, so are F f and F g . (Suggestion: Show that it makes sense to apply F to a chain homotopy s.) 3.10.4. Derived functors. Let us consider covariant, additive functors F from the category of R-modules to itself. (Recall that additive means that for any M, N , the induced map Hom(M, N ) → Hom(F (M ), F (N )) is a homomorphism of commutative groups.) Exercise: For any additive functor F and any chain complex C• of R-modules, F C• is again a chain complex. (Hint: the point here is that an additive functor takes the zero homomorphism to the zero homomorphism.) Thus if 0 −→ M1 −→ M2 −→ M3 −→ 0 is a short exact sequence of R-modules, then 0 −→ F (M1 ) −→ F (M2 ) −→ F (M3 ) −→ 0 is necessarily a complex of modules but not necessarily exact: it may have nonzero homology. Example: For any ring R, the functor F (M ) = M ⊕ M is exact. For R = Z the functor F (M ) = M ⊗ Z/2Z is not exact: for instance it takes the short exact sequence ·2 0 −→ Z −→ Z −→ Z/2Z −→ 0 to the complex ·2 0 −→ Z/2Z −→ Z/2Z → Z/2Z −→ 0, but multiplication by 2 on Z/2Z is not an injection. Although an exact functor is a thing of beauty and usefulness to all, it turns out that from a homological algebraic point of view, it is the functors which are “half exact” which are more interesting: they give rise to co/homology theories. 62 PETE L. CLARK An additive functor F is right exact if for any exact sequence of the form M1 −→ M2 −→ M3 −→ 0, the induced sequence F M1 −→ F M2 −→ F M3 −→ 0 is again exact. Note that this much was true for the functor F (M ) = M ⊗ Z/2Z, at least for the sequence we chose above. In fact this holds for all tensor products. Proposition 82. For any ring R and any R-module N , the functor F (M ) = M ⊗R N is right exact. Exercise: Prove Proposition 82. We have also the dual notion of an additive functor F being left exact: for any exact sequence of the form 0 → M1 → M2 → M3 , the induced sequence 0 → F M1 → F M2 → F M3 is again exact. We now wish to press our luck a bit by extending this definition to contravariant functors. Here a little abstraction actually makes me less confused, so I will pass it along to you: we say that a contravariant functor F from the abelian category C to the abelian category D is left exact (resp. right exact) if the associated covariant functor F opp : C opp → D is left exact (resp. right exact). Concretely, a contravariant functor F from R-modules to R-modules is left exact if every exact sequence of the form M1 → M2 → M3 → 0 is transformed to an exact sequence 0 → F M3 → F M2 → F M1 . (And similarly for right exact contravariant functors.) Proposition 83. Let R be a ring and X be an R-module. a) The functor M → Hom(X, M ) is covariant and left exact. (Recall that it is exact iff X if projective.) b) The functor M → Hom(M, X ) is contravariant and left exact. (Recall that it is exact iff X is injective.) Exercise X.X: Prove Proposition 83. Let F be a right exact additive functor on the category of R-modules. We will define a sequence {Ln F }n∈N of functors, with L0 F = F , called the left derived functors of F . The idea here is that the left-derived functors quantify the failure of F to be exact. Let M be an R-module. We define all the functors Ln M at once, as follows: COMMUTATIVE ALGEBRA 63 first we choose any projective resolution P• → M → 0 of M . Second we take away the M , getting a complex P• which is exact except at P0 , i.e., H0 (P ) = P0 / Im(P1 → P0 ) = P0 / Ker(P0 → M ) = M, ∀n > 0, Hn (P ) = 0. Third we apply the functor F getting a new complex F P• . And finally, we take homology of this new complex, defining (Ln F )(M ) := Hn (F P• ). Now there is (exactly?) one thing which is relatively clear at this point. Proposition 84. We have (L0 F )(M ) = F M . Proof. Since P1 → P0 → M → 0 is exact and F is right exact, F P1 → F P0 → F M → 0 is exact, hence Im(F P1 → F P0 ) = Ker(F P0 → F M ). Thus (L0 F )(M ) = H0 (F P• ) = Ker(F P0 → 0)/ Im(F P1 → F P0 ) = F P0 / Ker(F P0 → F M ) = F M. Before saying anything else about the left derived functors Ln F , there is an obvious point to be addressed: how do we know they are well-defined? On the face of it, they seem to depend upon the chosen projective resolution P• of M , which is very far from being unique. To address this point we need to bring in the Comparison ′ Theorem for Resolutions (Theorem 81). Namely, let P• → M → 0 be any other projective resolution of M . By Theorem 81, there exists a homomorphism of chain ′ complexes η : P• → P• which is unique up to chain homotopy. Interchanging the ′ ′ roles of P• and P• , we get a homomorphism η ′ : P• → P• . Moreover, the composition η ′ ◦ η is a homomorphism from P• to itself, so by the uniqueness η ′ ◦ η is chain homotopic to the identity map on P• . Similarly η ◦ η ′ is chain homotopic to the ′ identity map on P• , so that η is a chain homotopy equivalence. By Exercise X.X, ′ F η : F P• → F P• is a chain homotopy equivalence, and therefore the induced maps ′ on homology Hn (F η ) : Hn (F P• ) → Hn (F P• ) are isomorphisms. Thus we have shown that two different choices of projective resolutions for M lead to canonically isomorphic modules (Ln F )(M ). Exercise: Suppose M is projective. Show that for any right exact functor F and all n > 0, (Ln F )(M ) = 0. The next important result shows that a short exact sequence of R-modules induces a long exact sequence involving the left-derived functors and certain connecting homomorphisms (which we have not defined and will not define here). Theorem 85. Let (4) 0 −→ M1 −→ M2 −→ M3 −→ 0 64 PETE L. CLARK be a short exact sequence of R-modules, and let F be any left exact functor on the category of R-modules. Then: a) There is a long exact sequence (5) ∂ ∂ . . . → (L2 F )(M3 ) → (L1 F )(M1 ) → (L1 F )(M2 ) → (L1 F )(M3 ) → F M1 → F M2 → F M3 → 0. b) The above construction is functorial in the following sense: if 0 −→ N1 −→ N2 −→ N3 −→ 0 is another short exact sequence of R-modules and we have maps Mi → Ni making a “short commutative ladder”, then there is an induced “long commutative latter” with top row the long exact sequence associated to the first short exact sequence and the bottom row the long exact sequence associated to the second short exact sequence. Proof. No way. See [W, Thm. 2.4.6]. Remark: One says that (5) is the long exact homology sequence associated to the short exact sequence (4). Now, dually, if F is a right exact functor on the category of R-modules, we may define right derived functors Rn F . Namely, for an R-module M , first choose an injective resolution 0 → M → E • , then take M away to get a cochain complex E • , then apply F to get a cochain complex F E • , and then finally define (Rn F )(M ) = H n (F E • ). In this case, a short exact sequence of modules (4) induces a long exact cohomology sequence (6) ∂ ∂ 0 → F M1 → F M2 → F M3 → (R1 F )(M1 ) → (R1 F )(M2 ) → (R1 F )(M3 ) → (R2 F )(M1 ) . . . Exercise: Suppose M is injective. Show that for any left exact functor F and all n > 0, (Rn F )(M ) = 0. 3.10.5. Tor. Let M, N be R-modules, and let F : N → M ⊗R N be the functor “tensor with M ”. By X.X F is right exact so has left derived functors (Ln F ). By definition, for all n ∈ N, Torn (M, N ) := (Ln F )(N ). Now un/fortunately the situation is even a little richer than the general case of left-derived functors discussed above. Namely, the tensor product is really a bifunctor: i.e., a functor in M as well as in N , additive and covariant in each variable separately. So suppose we took the right-derived functors of M → M ⊗R N and applied them to M : this would give us Torn (N, M ). So it is natural to ask: how does Torn (M, N ) compare to Torn (N, M )? Since for n = 0 we have that M ⊗R N is canonically isomorphic to N ⊗R M , it is natural to hope that the Tor functors are symmetric. And indeed this turns out to be the case. Theorem 86. (Balancing Tor) For any R-modules M and N and all n ≥ 0, there are natural isomorphisms Torn (M, N ) = Torn (N, M ). Proof. No way. See [W, Thm. 2.7.2]. Exercise: In order to use the Universal Coefficient Theorem (for homology) in algebraic topology, it is necessary to know the values of Tor1 (M, N ) for any two finitely COMMUTATIVE ALGEBRA 65 generated Z-modules M and N . a) Show that for any m, n ∈ Z+ , Tor1 (Z/mZ, Z/nZ) ∼ Z/ gcd(m, n)Z. = b) Show that for all Z-modules N , Tor1 (Z, N ) = 0. c) Explain how the structure theorem for finitely generated Z-modules reduces the problem of computation of Tor1 (M, N ) for any finitely generated M and N to the two special cases done in parts a) and b). Exercise: Show that the tor functors commute with direct limits: for all n ∈ N, any directed system {Mi }i∈I of R-modules M and any R-module N we have a canonical isomorphism Torn (lim Mi , N ) → lim Torn (Mi , N ). − → − → i i (Suggestion: the case n = 0 is Proposition 17. Use this to show the general case by brute force: i.e., take a projective resolution of N and track these isomorphisms through the definition of Torn .) 3.10.6. Ext. Nota Bene: At the present time, we do not use the Ext functors for anything in these notes. However they certainly do appear in commutative algebra and elsewhere. Moreover, having taken the trouble (and it was some trouble!) to set up enough machinery to define the Tor functors, we might as well follow it up with the parallel disucssion of the Ext functors. Let M, N be R-modules, and let F : N → Hom(M, N ). By Proposition X.X, F is covariant and left exact. By definition, for all n ∈ N, Extn (M, N ) = (Rn F )(N ). But again, we have an embarrassment of riches: why didn’t we define the Ext functors as the right-derived functors of the contravariant left exact functor G : N → Hom(N, M )? Again, we can do this. Theorem 87. (Balancing Ext) Let M and N be R-modules. Define functors FM : N → Hom(M, N ) and GN : M → Hom(M, N ). Then for all n ≥ 0, (Rn FM )(N ) = (Rn GN )(M ). Proof. No way. See [W, Thm. 2.7.6]. Exercise: In order to use the Universal Coefficient Theorem (for cohomology) in algebraic topology, it is necessary to know the values of Ext1 (M, N ) for any two finitely generated Z-modules M and N . Compute them. (Hint: as for the analgous problem with Tor, it is enough to consider four special cases: (M, N ) = (Z, Z), (M, N ) = (Z, Z/nZ), (M, N ) = (Z/mZ, Z), (M, N ) = (Z/mZ, Z/nZ).) 3.11. More on flat modules. Theorem 88. (Tensorial Criterion for Flatness) For an R-module M , TFAE: (i) M is flat. (ii) For every finitely generated ideal I of R the canonical map I ⊗R M → IM is an isomorphism. 66 PETE L. CLARK Proof. Fist note that the canonical map I ⊗R M → IM is always a surjection. (i) =⇒ (ii): if M is flat, then since I → R, I ⊗R M → R ⊗R M = M , so ∼ I ⊗R M → IM . (ii) =⇒ (i): Every ideal of R is the direct limit of its finitely generated subideals, so it follows from Proposition 17 and the exactness of direct limits that I ⊗ M → M is injective for all ideals I . Moreover, if N is an R-module and N ′ ⊂ N is an Rsubmodule, then since N is the direct limit of submodule N ′ + F with F finitely generated, to show that N ′ ⊗ M → N ⊗ M is injective we may assume N = N ′ + ⟨ω1 , . . . , ωn ⟩R . We now proceed by d´vissage: putting Ni = N ′ + ⟨ω1 , . . . , ωi ⟩R , it is enough to e show injectivity at each step of the chain N ′ ⊗ M → N1 ⊗ M → . . . → N ⊗ M, and further simplifying, it is enough to show that if N = N ′ + Rω , then N ′ ⊗ M → N ⊗ M . Let I be the “conductor ideal of N/N ′ ”, i.e., I = {x ∈ R | xω ∈ N ′ }, so that we get a short exact sequence of R-modules 0 → N ′ → N → R/I → 0 which gives rise to a long exact homology sequence . . . → TorR (M, R/I ) → N ′ ⊗ M → N ⊗ M → M/IM → 0. 1 Thus it suffices to prove TorR (M, R/I ) = 0. For this we consider the homology 1 sequence associated to 0 → I → R → R/I → 0, namely . . . → TorR (M, R) = 0 → TorR (M/R/I ) → I ⊗ M → M → . . . , 1 1 and from the injectivity of I ⊗ M → M we deduce TorR (M, R/I ) = 0. 1 Theorem 89. (Homological Criterion for Flatness) For a module M over a ring R, TFAE: (i) M is flat. (ii) For every R-module N all i > 0, TorR (M, N ) = 0. i (ii′ ) For every R-module N , TorR (M, N ) = 0. 1 (iii) For every finitely generated ideal I of R, TorR (M, R/I ) = 0. 1 Proof. (i) =⇒ (ii): This is a statement about projective resolutions, but given that it is just about the most basic possible one. Namely, let L• → N → 0 be a projective resolution of N . Then . . . → Ln ⊗ M → Ln−1 ⊗ M → . . . → L0 ⊗ M is exact, so TorR (M, N ) = 0 for all i > 0. i (ii) =⇒ (ii′ ) and (ii′ ) =⇒ (iii) are both immediate. (iii) =⇒ (i): For each finitely generated ideal I of R, the short exat sequence 0 → I → R → R/I → 0 of R-modules induces a long exact sequence in homology, which ends . . . → TorR (M, R/I ) = 0 → I ⊗ M → M → M/IM → 0, 1 COMMUTATIVE ALGEBRA 67 ∼ i.e., the map I ⊗ M → M is injective and thus induces an isomorphism I ⊗ M → IM . Using the Tensorial Criterion for Flatness (Theorem 88), we conclude M is flat. Corollary 90. (Direct limits preserve flatness) Let R be a ring and {Mi }i∈I a directed system of flat R-modules. Then M = lim Mi is a flat R-module. − → Proof. For every R-module N , we have TorR (lim Mi , N ) ∼ TorR (N, lim Mi ) = lim TorR (N, Mi ) ∼ lim TorR (Mi , N ) = lim 0 = 0. = =− 1− 1 1 1 → − → − → → − → Now apply the Homological Criterion for Flatness. Corollary 91. For a domain R, TFAE: (i) Every finitely generated torsionfree R-module is flat. (ii) Every torsionfree R-module is flat. Proof. Every submodule of a torsionfree R-module is torsionfree, and every Rmodule is the direct limit of its finitely generated submodules. So the result follows immediately from Proposition 90. Corollary 92. Let 0 → M ′ → M → M ′′ → 0 be a short exact sequence of Rmodules, with M ′′ flat. Then M ′ is flat iff M is flat. Exercise X.X: Use the Homological Criterion of Flatness to prove Corollary 92. Exercise X.X: In a short exact sequence of R-modules as in Corollary 92, if M ′ and M are flat, must M ′′ be flat? Now recall that a finitely generated torsion free module over a PID is free (Proposition 66). From this we deduce: Corollary 93. A module over a PID is flat iff it is torsionfree. Exercise: Let R be a domain and M a torsion R-module. Show that for all Rmodules N and all n ≥ 0, Torn (M, N ) is a torsion R-module. Theorem 94. Let R be a PID and let M, N be R-modules. a) For all n ≥ 2, Torn (M, N ) = 0. b) Tor1 (M, N ) is a torsion R-module. Proof. a) Choose a free module F0 and a surjection d0 : F0 → N . By X.X, F1 = Ker(d0 ) is free, so we get a finite free resolution of N: 0 → F1 → F0 → N → 0. Therefore we certainly have Torn (M, N ) for all M and all n ≥ 2. b) Let {Mi }i∈I be the direct system of all finitely generated submodules of M . As above, we have M = lim Mi , so − → Tor1 (M, N ) = Tor1 (lim Mi , N )) = lim Tor1 (Mi , N ). − → − → By Corollary 93, each Mi which is torsionfree is flat, hence Tor1 (Mi , N ) = 0. Thus the only possible contribution to lim Tor1 (Mi , N ) comes from torsion modules Mi , − → and by Exercise X.X, Mi torsion implies Tor1 (Mi , N ) torsion. Thus Tor1 (M, N ) is a direct limit of torsion modules, hence itself a torsion module. 68 PETE L. CLARK Theorem 95. (Equational Criterion for Flatness) Let M be an R-module. a) Suppose M is flat, and that we are given r, n ∈ Z+ , a matrix A = (aij ) ∈ Mr×n (R) and elements x1 , ldots, xn ∈ M such that ∑ ∀1 ≤ i ≤ r, aij xj = 0. j Then there exists s ∈ Z+ , bjk ∈ R and yk ∈ M (for 1 ≤ j ≤ n and 1 ≤ k ≤ s) such that ∑ ∀i, k, aij bj = 0 j and ∀j, xj = ∑ bjk yk = 0. j Thus the solutions in a flat module of a system of linear equations with R-coefficients can be expressed as a linear combination of solutions of the system in R. b) Conversely, if the above conditions hold for a single equation (i.e., with r = 1), then M is a flat R-module. Proof. a) Let φ : Rn → Rr be the linear map corresponding to multiplication by the matrix A and let φM : M n → M r be the same for M , so that φM = φ ⊗ 1M . Let K = Ker φ. Since M is flat, tensoring with M preserves exact sequences, thus the sequence ι⊗1 φ K ⊗R M → M n → M r is exact. By our hypothesis we have φM (x1 , . . . , xn ) = 0, so that we may write ) (s ∑ βk ⊗ yk (x1 , . . . , xn ) = (ι ⊗ 1) k=1 with βk ∈ K and yk ∈ M . Writing out each βk as an element (b1k , . . . , bnk ) ∈ Rn gives the desired conclusion. b) We will use the Tensorial Criterion for Flatness to show that M is flat. Let I = ⟨a1 , . . . , an ⟩ be a ∑ finitely generated ideal of R. We may write an arbitrary ∑n n element z of I ⊗ M as i=1 ai ⊗ mi with mi ∈ M . Let z = i=1 ai mi denote the image of z in IM ⊂ M . We want to show that z = 0 implies z = 0, so suppose ∑ that i ai mi = 0. By hypothesis, there exist bij ∈ R and yj ∈ M such that for all ∑ ∑ j , i ai bij = 0 and for all i, mi = j bij yj . Thus ( ) ∑ ∑∑ ∑∑ ∑ z= ai ⊗ mi = ai bij ⊗ yj = ai bij ⊗ yj = 0 ⊗ yj = 0. i i j j i j As an application, we can now improve Theorem 58 by weakening the hypothesis of “finite presentation” to the simpler one of “finite generation”. Theorem 96. Let M be a finitely generated flat module over the local ring (R, m). Then for all n ∈ Z+ , x1 , . . . , xn are elements of M such that the images in R/m are R/m-linearly independent, then x1 , . . . , xn are R-linearly independent. COMMUTATIVE ALGEBRA 69 Proof. We go by induction on n. Suppose first that n = 1, in which case it is sufficient to show that a1 ∈ R, a1 x1 ̸= 0 implies a1 = 0. By the Equational Criterion for Flatness, there exist b1 , . . . , bs ∈ R such that abi = 0 for all i and ∑ x1 ∈ i bi M . By assumption, x1 does not lie in mM , so that for some i we must have bi ∈ R× , and then abi = 0 implies a = 0. Now suppose n > 1, and let a1 , . . . , an ∈ R are such that a1 x1 + . . . + an xn = 0. Again using the Equational Criterion for Flatness, there are bij ∈ R and y1 , . . . , ys ∈ ∑ ∑ M such that for all j , ai bij = 0 and xi = bij yj . Since the set of generators is minimal, by Nakayama’s Lemma their images in M/mM must be R/m-linearly independent. In particular xn ∈ mM , so that at least one bnj is a unit. It follows / ∑n−1 that there exist c1 , . . . , cn−1 ∈ R such that an = i=1 ai ci . Therefore a1 (x1 + c1 xn ) + . . . + an−1 (xn−1 + cn−1 xn ) = 0. The images in M/mM of the n − 1 elements x1 + c1 xn , . . . , xn−1 + cn−1 xn are R/m-linearly independent, so by induction a1 = . . . = an−1 = 0. Thus an = 0. Theorem 97. For a finitely generated module M over a local ring R, TFAE: (i) M is free. (ii) M is projective. (iii) M is flat. Proof. For any module over any ring we have (i) =⇒ (ii) =⇒ (iii). So suppose that M is a finitely generated flat module over the local ring (R, m). Let (x1 , . . . , xn ) be a set of R-module generators for M of minimal cardinality. By Nakayama’s Lemma the images of x1 , . . . , xn in R/m are R/m-linearly independent, and then Theorem 96 implies that x1 , . . . , xn is a basis for M as an R-module. A ring R is called absolutely flat if every R-module is flat. Exercise: Show that any quotient of an absolutely flat ring is absolutely flat. Proposition 98. For a ring R, TFAE: (i) R is absolutely flat. (ii) For every principal ideal I of R, I 2 = I . (iii) Every finitely generated ideal of R is a direct summand of R. Proof. (i) =⇒ (ii): Assume R is absolutely flat, and let I = (x) be a principal ideal. Tensoring the natural inclusion (x) → R with R/(x), we get a an injection (x) ⊗R R/(x) → R/(x). But this map sends x ⊗ r → xr + (x) = (x), so it is identically zero. Therefore its injectivity implies that 0 = (x) ⊗R R/(x) ∼ (x)/(x2 ), = so (x) = (x2 ). (ii) =⇒ (iii): Let x ∈ R. Then x = ax2 for some a ∈ R, so putting e = ax we have e2 = a2 x2 = a(ax2 ) = ax = e, so e is idempotent, and (e) = (x). In general, for any two idempotents e, f , we have ⟨e, f ⟩ = (e + f − ef ). Hence every finitely generated ideal is principal, generated by an idempotent element, and thus a direct summand. (iii) =⇒ (i): Let M be an R-module, and let I be any finitely generated ideal of R. By assumption, we may choose J such that R = I ⊕ J . Therefore J is projective, so Tor1 (R/I, M ) = Tor1 (J, M ) = 0. By the Homological Criterion for Flatness, M is flat. 70 PETE L. CLARK The following striking result came relatively late in the game: it is due independently to Govorov [Gov65] and Lazard [Laz64]. Theorem 99. (Govorov-Lazard) For a module M over a ring R, TFAE: (i) M is flat. (ii) There exists a directed family {Fi }i∈I of finitely generated free submodules of M such that M = lim Fi . − → Proof. (i) =⇒ (ii): Suppose M = lim Fi is a direct limit of free modules. Then in − → particular M is a direct limit of flat modules, so by Corollary 90 M is flat. (ii) =⇒ (i): see [Eis, Thm. A6.6]. 3.11.1. Flat Base Change. Proposition 100. (Stability of flatness under base change) Let M be a flat Rmodule, and f : R → S a ring homomorphism. Then S ⊗R M is a flat S -module. Exercise X.X: Prove Proposition 100. Exercise X.X: Show that the tensor product of flat R-modules is a flat R-module. Exercise X.X: Let R be a nonzero commutative ring, and n, m ∈ N. a) Show that Rm ∼ Rn iff m = n. = b) Suppose that φ : Rm → Rn is a surjective R-module map. Show that m ≥ n. c)18 Suppose that φ : Rm → Rn is an injective R-module map. Show that m ≤ n. d) Find a noncommutative ring R for which part a) fails. Theorem 101. (Hom commutes with flat base change) Let S be a flat R-algebra and M, N R-modules with M finitely presented. Then the canonical map ΦM : S ⊗R HomR (M, N ) → HomS (M ⊗R S, N ⊗R S ) induced by (s, f ) → (m ⊗ t) → f (m) ⊗ st is an isomorphism. Proof. (Hochster) It is immediate that ΦR is an isomorphism and that ΦM1 ⊕M2 = ΦM1 ⊕ ΦM2 , and thus ΦM is an isomorphism when M is finitely generated free. For finitely presented M , there is an exact sequence H→G→M →0 with H and G finitely generated free modules. Now we have the following commutative diagram: 0 −→ 0 θM S ⊗R HomR (M, N ) −→ HomS (M ⊗R S, N ⊗R S ) θ G S ⊗R HomR (G, N ) −→ HomS (G ⊗R S, N ⊗R S ) θ H S ⊗R HomR (H, N ) −→ HomS (H ⊗R S, N ⊗R S ). Note that the right column is obtained by first applying the exact functor A → A ⊗R S and then applying the right exact cofunctor U → HomS (U, N ⊗R S ), so it is exact. Similarly, the left column is obtained by first applying the right exact cofunctor A → HomR (A, N ) and then applying the exact (since R is flat) functor A → A ⊗R S , so is exact. Since G and H are finitely generated free, θG and θH are isomorphisms, and a diagram chase shows that θM is an isomorphism. 18This is actually quite challenging. COMMUTATIVE ALGEBRA 71 3.12. Faithful flatness. Proposition 102. For an R-module M , TFAE: (i) For a sequence (7) α β N1 −→ N2 −→ N3 of left R-modules to be exact it is necessary and sufficient that (8) A B M ⊗R N1 −→ M ⊗R N2 −→ M ⊗R N3 be exact. (ii) M is flat and for all nonzero R-modules N , M ⊗R N ̸= 0. (iii) M is flat and for all nonzero R-module maps u : N → N ′ , 1M ⊗ u : M ⊗R N → M ⊗R N ′ is not zero . (iv) M is flat and for every m ∈ MaxSpec R, mM M . (v) M is flat and for every p ∈ Spec R, pM M . A module satisfying these equivalent conditions is faithfully flat. Proof. (i) =⇒ (ii): Certainly (i) implies that M is flat. Moreover, if N is a nonzero R-module such that M ⊗ N = 0, then 0 → N → 0 is not exact but its tensor product with M is exact, contradicting (i). (ii) =⇒ (iii): Let I = Im(u); then M ⊗ I = Im(1M ⊗ u). So assuming (ii) and that I ̸= 0, we conclude Im(1M ⊗ u) ̸= 0. (iii) =⇒ (i): Assume (iii). Then, since M is flat, if (7) is exact, so is (8). Conversely, suppose (8) is exact, and put I = Im(α), K = ker(β ). Then B ◦ A = 1M ⊗ (β ◦ α) = 0, so β ◦ α = 0, or in other words, I ⊂ K . We may therefore form the exact sequence 0 → I → K → K/I → 0, and tensoring with the flat module M gives an exact sequence 0 → M ⊗ I → M ⊗ K → M ⊗ K/I → 0. But M ⊗ K = M ⊗ I by hypothesis, so K/I = 0 and I = K . (ii) =⇒ (iv): Let m ∈ MaxSpec R. Then R/m is a nonzero R-module, so by (ii) so is M ⊗ R/m = M/mM , i.e., mM M . (iv) =⇒ (ii): Assume (iv) holds. Then, since every proper ideal is contained in a maximal ideal, we have moreover that for all proper ideals I of R, IM M , or equivalently M ⊗ (R/I ) ̸= 0. But the modules of the form R/I as I ranges over all proper ideals of R are precisely all the cyclic (a.k.a. monogenic) R-modules, up to isomorphism. Now if N is any nonzero R-module, choose 0 ̸= x ∈ M and let N ′ = ⟨x⟩ by the cyclic submodule spanned by x. It follows that M ⊗ N ′ ̸= 0. Since M is flat, N ′ → N implies M ⊗ N ′ → M ⊗ N , so M ⊗ N ̸= 0. (iv) ⇐⇒ (v): this follows immediately from the proofs of the last two implications, as we leave it to the reader to check. Exercise: Show that (iv) ⇐⇒ (v) in Proposition 102. Corollary 103. Let M be a faithfully flat and u : N → N ′ an R-module map. Then: a) u is injective iff 1M ⊗ u : M ⊗ N → M ′ N is injective. b) u is surjective iff 1M ⊗ u is surjective. c) u is an isomorphism iff 1M ⊗ u is an isomorphism. 72 PETE L. CLARK Exercise: Deduce Corollary 103 from Proposition 102. Exercise X.X: Use each of the criteria of Proposition 102 to show that the (flat) Z-module Q is not faithfully flat. Exercise X.X: Show that a faithfully flat module is faithful and flat, and that – unfortunately! – a flat, faithful module need not be faithfully flat. Exercise X.X: Show that a nonzero free module is faithfully flat but that a nonzero (even finitely generated) projective module need not be. ⊕ Exercise X.X: Let {Mi }i∈I be a family of flat R-modules, and put M = i∈I Mi . a) Suppose that for some i, Mi is faithfully flat. Show that M is faithfully flat. b) Give an example where no Mi is faithfully flat yet M is faithfully flat. Proposition 104. Let 0 → M ′ → M → M ′′ → 0 be a short exact sequence of R-modules. Suppose M ′ and M ′′ are flat and that at least one is faithfully flat. Then M is faithfully flat. Proof. By Proposition 92, M is flat. Now let N be any R-module. Since M ′′ is flat, Tor1 (M ′′ , N ) = 0 so 0 → M ′ ⊗ N → M ⊗ N → M ′′ ⊗ N → 0 is exact. Thus if M ⊗ N = 0 then M ′ ⊗ N = M ′′ ⊗ N = 0. Since one of M ′ , M ′′ is faithfully flat, by criterion (ii) of Proposition 102 we have N = 0, and then that same criterion shows that M is faithfully flat. By a faithfully flat R-algebra, we mean a ring S equipped with a ring homomorphism R → S making S into a faithfully flat R-module. Proposition 105. Let f : R → S be a ring map and M an R-module. Then: M is faithfully flat iff M ⊗R S is faithfully flat. Proof. The key fact is that for any S -module N , we have (M ⊗R S ) ⊗S N ∼R M ⊗R N. = With this, the proof becomes straightforward and is left to the reader. Exercise: Complete the proof of Proposition 105. Theorem 106. For a flat algebra f : R → S , TFAE: (i) S is faithfully flat over R. (ii) f ∗ : MaxSpec S → MaxSpec R is surjective. (iii) f ∗ : Spec S → Spec R is surjective. Proof. (i) ⇐⇒ (ii): Let m be any maximal ideal of R. Then mS S holds iff there is a maximal ideal M of S containing mS iff f ∗ (M) = m. The equivalence now follows from criterion (iv) of Proposition 102. (i) =⇒ (iii): Let p ∈ Spec R, and let k (p) be the fraction field of the domain R/p. By faithful flatness, S ⊗R k (p) is a nonzero k (p)-algebra so has a prime f g ideal P . Consider the composite map h : R → S → S ⊗R k (p). We claim that g ∗ : Spec(S ⊗R k (p)) → Spec R has image precisely {p}. The proof of this result, a spectral description of the fiber of the morphism f : R → S over p, will have to COMMUTATIVE ALGEBRA 73 wait until we have developed the theory of localization in §7.3. Assuming it for now, we get that g ∗ (P ) is a prime ideal of Spec S such that f ∗ g ∗ (P ) = (g ◦ f )∗ (P ) = p, so f ∗ : Spec S → Spec R is surjective. (iii) =⇒ (ii): Let m ∈ MaxSpec R ⊂ Spec R. By assumption, the set of prime ideals P of S such that f ∗ P = m is nonempty. Moreover the union of any chain of prime ideals pulling back to m is again a prime ideal pulling back to p, so by Zorn’s Lemma there exists an ideal M which is maximal with respect to the property that f ∗ M = m. Suppose M is not maximal and let M′ be a maximal ideal properly containing M. Then by construction f ∗ (M′ ) properly contains the maximal ideal m of R, i.e., f ∗ (M)′ ) = R, contradicting the fact that prime ideals pull back to prime ideals. So M is indeed maximal in S . Proposition 107. Let f : R → S be a ring extension such that S is a faithfully flat R-module, and let M be an R-module. Then: a) M is finitely generated iff M ⊗R S is finitely generated. b) M is finitely presented iff M ⊗R S is finitely presented. Proof. Note first that the properties of finite generation and finite presentation are preserved by arbitrary base change f : R → S . So it suffices to prove that if M ⊗R S is finitely generated (resp. finitely presented), then M is finitely generated (resp. finitely presented). a) Since M ⊗R S is finitely generated over S , it has a finite set of S -module generators of the form xi ⊗ 1. Let N = ⟨x1 , . . . , xn ⟩R and ι : N → M the canonical injection. Then ιS : N ⊗R S → M ⊗R S is an isomorphism, so by faithful flatness ι was itself an isomorphism and thus M = ⟨x1 , . . . , xn ⟩ is finitely generated. b) By part a), M is finitely generated over R, so let u : Rn → M be a surjection. Since M ⊗R S is finitely presented, the kernel of uS : S n → M ⊗R S is finitely generated over S . Since by flatness ker uS = (ker u)S , part a) shows that ker u is finitely generated and thus that M is finitely presented. Lemma 108. Let f : R → S be a ring map, and let M, N be R-modules. a) There is a canonical S -module map ω : HomR (M, N ) ⊗R S → HomS (M ⊗R S, N ⊗R S ) such that for all u ∈ HomR (M, N ), ω (u ⊗ 1) = u ⊗ 1B . b) If S is flat over R and M is finitely generated, then ω is injective. c) If S is flat over R and M is finitely presented, then ω is an isomorphism. Exercise: Prove Lemma ??. (It is not difficult, really, but it is somewhat technical. Feel free to consult [?, p. 23] for the details.) Theorem 109. (Faithfully flat descent for projective modules) Let f : R → S be a faithfully flat ring extension, and let P be an R-module. Then P is finitely generated and projective iff P ⊗R S is finitely generated and projective. Proof. Begin, once again the implication P finitely generated projective implies P ⊗R S is finitely generated projective holds for any base change. So suppose P ⊗R S is finitely generated projective. Then P ⊗R S is finitely presented, so by Proposition 107, M is finitely presented. It remains to show that M is projective. Let v : M → N be a surjection of R-modules. We wish to show that the natural map HomR (P, M ) → HomR (P, N ) is surjective. Because of the faithful flatness 74 PETE L. CLARK of S/R, it is sufficient to show that HomR (P, M ) ⊗R S → HomR (P, N ) ⊗R S is surjective, and by Lemma 108 this holds iff HomS (P ⊗R S, M ⊗R S ) → HomS (P ⊗R S, N ⊗R S ) is surjective. But this latter map is surjective because M ⊗R S → N ⊗R S is surjective and the S -module P ⊗R S is projective by assumption. 4. First Properties of Ideals in a Commutative Ring 4.1. Introducing maximal and prime ideals. Consider again the set I (R) of all ideals of R, partially ordered by inclusion. The maximal element is the ideal R itself, and the minimal element is the ideal (0). In general our attitude to the ideal R of R is as follows: although we must grudgingly admit its existence – otherwise, given a subset S of R it would be in general a difficult question to tell whether the ideal ⟨S ⟩ generated by S “exists” (i.e., is proper) or not – nevertheless we regard it as exceptional and try to ignore it as much as possible. Because of this we define an ideal I of R to be maximal if it is maximal among all proper ideals of R, i.e., I R and there does not exist J such that I J R. That this is a more interesting concept than the literally maximal ideal R of R is indicated by the following result. Proposition 110. For an ideal I of R, TFAE: (i) I is maximal. (ii) R/I is a field. Proof. Indeed, R/I is a field iff it has precisely two ideals, I and R, which by the Correspondence Theorem says precisely that there is no proper ideal strictly containing I . Example: In R = Z, the maximal ideals are those of the form (p) for p a prime number. The quotient Z/pZ is the finite field of order p. Does every ring have a maximal ideal? With a single (trivial) exception, the answer is yes, assuming – as we must, in order to develop the theory as it is used in other branches of mathematics – suitable transfinite tools. Proposition 111. Let R be a nonzero ring and I a proper ideal of R. Then there exists a maximal ideal of R containing I . Proof. Consider the set S of all proper ideals of R containing I , partially ordered by inclusion. Since I ∈ S , S is nonempty. Moreover the union of a chain of ideals is an ideal, and the union of a chain of proper ideals is proper (for if 1 were in the union, it would have to lie in one of the ideals of the chain). Therefore by Zorn’s Lemma we are entitled to a maximal element of S , which is indeed a maximal ideal of R that contains I . Corollary 112. A nonzero ring R contains at least one maximal ideal. Proof. Apply Proposition 111 with I = (0). COMMUTATIVE ALGEBRA 75 Remark: The zero ring has the disquieting property of having no maximal ideals. Remark: The appeal to Zorn’s Lemma cannot be avoided, in the sense that Corollary 112 implies the Axiom of Choice. A proper ideal I of a ring R is prime if xy ∈ I implies x ∈ I or y ∈ I . Exercise 4.1: Let p be a prime ideal of R. a) Suppose x1 , . . . , xn are elements of R such that x1 · · · xn ∈ p. Then xi ∈ p for some at least one i. b) In particular, for x ∈ R and n ∈ Z+ we have xn ∈ p, then x ∈ p. Proposition 113. Let f : R → S be a homomorphism of rings, and let J be an ideal of S . a) Put f ∗ (J ) := f −1 (J ) = {x ∈ R | f (x) ∈ J }. Then f ∗ (J ) is an ideal of R. b) If J is a prime ideal, so is f ∗ (J ). Exercise 4.2: Prove Proposition 113. Proposition 114. For a commutative ring R, TFAE: (i) If x, y ∈ R are such that xy = 0, then x = 0 or y = 0. (ii) If 0 ̸= x ∈ R and y, z ∈ R are such that xy = xz , then y = z . A ring satisfying either of these two properties is called an integral domain. Proof. Assume (i), and consider xy = xz with x ̸= 0. We have x(y − z ) = 0, and since x ̸= 0, (i) implies y − z = 0, i.e., y = z . Assuming (ii) suppose xy = 0 with x ̸= 0. Then xy = 0 = x · 0, so applying cancellation we get y = 0. A zero divisor in a ring R is an element x such that there exists 0 ̸= y ∈ R with xy = 0. (In particular 0 is a zero divisor, albeit not a very interesting one.) So property (i) expresses that there are no zero divisors other than 0 itself. Property (ii) makes sense in any commutative monoid and is called cancellation. Remark: The terminology “integral domain” is motivated by the fact the integers Z satisfy (i) and (ii) of Proposition 114, so any ring which satisfies these properties can be viewed as a sort of ring of “generalized integers.” The analogy is apt – in particular, later on we shall build from any integral domain a field of fractions in exactly the same way that the rational numbers are constructed from the integers – but the terminology is quite awkward, as the reader will come to appreciate. First of all the word “integral” is logically superfluous – we do not have any other definition of a domain, and indeed often we will use domain as a more succinct synonym for “integral domain”. So why not just “domain”? One problem is that, being a property of an object and not an object itself, we would prefer an English word which is an adjective rather than a noun. So why not just “integral”? The problem is that the word “integral” will be used later for something else (not a property of a single ring but a property of an extension ring S of R). It would certainly be confusing to use the term integral for these two different concepts. Ideal would probably be to reserve the term “integral” for a ring without zero divisors, give some other name to integral extensions (integrally algebraic?), and eliminate the use of “domain” from terms like “principal ideal domain.” (This would be consistent with geometric terminology: an affine scheme Spec R is called integral 76 PETE L. CLARK iff R is an integral domain.) However, the terminology is too entrenched for this to be a feasible solution. Proposition 115. For an ideal I in a ring R, TFAE: (i) I is prime. (ii) R/I is an integral domain. Exercise 4.3: Prove Proposition 115. Corollary 116. A maximal ideal is prime. Proof. If I is maximal, R/I is a field, hence an integral domain, so I is prime. Corollary 116 is the first instance of a somewhat mysterious meta-principle in ideal theory: for some property P of ideals in a ring, it is very often the case that an ideal which is maximal with respect to the satisfaction of property P (i.e., is not strictly contained in any other ideal satisfying P) must be prime. In the above, we saw this with P = “proper”. Here is another instance: Proposition 117. (Multiplicative Avoidance) Let R be a ring and S ⊂ R. Suppose: 1 is in S ; 0 is not in S ; and S is closed under multiplication: S · S ⊂ S . Let IS be the set of ideals of R which are disjoint from S . Then: a) IS is nonempty;. b) Every element of IS is contained in a maximal element of IS . c) Every maximal element of IS is prime. Proof. a) (0) ∈ IS . b) Let I ∈ IS . Consider the subposet PI of IS consisting of ideals which contain I . Since I ∈ PI , PI is nonempty; moreover, any chain in PI has an upper bound, namely the union of all of its elements. Therefore by Zorn’s Lemma, PI has a maximal element, which is clearly also a maximal element of IS . c) Let I be a maximal element of IS ; suppose that x, y ∈ R are such that xy ∈ I . If x is not in I , then ⟨I, x⟩ I and therefore contains an element s1 of S , say s1 = i1 + ax. Similarly, if y is not in I , then we get an element s2 of S of the form s2 = i2 + by. But then s1 s2 = i1 i2 + (by )i1 + (ax)i2 + (ab)xy ∈ I ∩ S, a contradiction. In fact Corollary 116 is precisely the special case S = {1} of Proposition 117. If I and J are ideals of R, we define the product IJ to be the ideal generated by all∑ elements of the form xy with x ∈ I, y ∈ J . Every element of IJ is of the n form i=1 xi yi with x1 , . . . , xn ∈ I, y1 , . . . , yn ∈ J . The following simple result will be used many times in the sequel. Proposition 118. Let p be a prime ideal and I1 , . . . , In be ideals of a ring R. If p ⊃ I1 · · · In , then p ⊃ Ii for at least one i. Proof. An easy induction argument reduces us to the case of n = 2. So suppose for a contradiction that p ⊃ I1 I2 but there exists x ∈ I1 \ p and y ∈ I2 \ p. Then xy ∈ I1 I2 ⊂ p; since p is prime we must have x ∈ p or y ∈ p, contradiction. COMMUTATIVE ALGEBRA 77 Exercise 4.4: Show that Proposition 118 characterizes prime ideals, in the sense that if p is any ideal such that for all ideals I, J of R, p ⊂ IJ implies p ⊂ I or p ⊂ JJ , then p is a prime ideal. For an ideal I and n ∈ Z+ , we denote the n-fold product of I with itself by I n . Corollary 119. If p is a prime ideal and I is any ideal, then p ⊃ I n =⇒ p ⊃ I . 4.2. Radicals. An element x of a ring R is nilpotent if xn = 0 for some n ∈ Z+ . Obviously 0 is a nilpotent element; a ring in which 0 is the only nilpotent element is called reduced. An ideal I of R is nil if every element of I is nilpotent. An ideal I is nilpotent if there exists n ∈ Z+ such that I n = (0). Proposition 120. Let I be an ideal of a ring R. a) If I is nilpotent, then I is a nil ideal. b) If I is finitely generated and nil, then I is nilpotent. Proof. Part a) is immediate from the definition, as we invite the reader to check. Suppose I = ⟨a1 , . . . , an ⟩R . Since I is nil, for each i, 1 ≤ i ≤ r, there exists ni such that ani = 0. Let N = n1 + . . . + nr . We claim I N = 0. Indeed, an arbirary i element of I is of the form x1 a1 + . . . + xn an . Raising this element to the nth power ∑r yields a sum of monomials of the form xj1 · · · xjr aj1 · · · ajr , where i=1 ji = N . If r r1 1 we had for all i that ji < ni , then certainly j1 + . . . + jr < N . So for at least one i we have ji ≥ ni and thus xji = 0; so every monomial term equals zero. i Exercise 4.5: Find a ring R and an ideal I of R which is nil but not nilpotent. The nilradical N of R is the set of all nilpotent elements of R. Proposition 121. Let R be a ring. a) The nilradical N is a nil ideal of R. b) The quotient R/N is reduced. c) The map q : R → R/N is universal for maps from R into a reduced ring. d) The nilradical is the intersection of all prime ideals of R. Proof. a) In establishing that N is an ideal, the only property which is not absolutely immediate is its closure under addition. Suppose xm = 0 = y n . Then every term in the binomial expansion of (x + y )m+n−1 is an integer times xi y m+n−1−i for 0 ≤ i ≤ m + n. Here either i ≥ m so xi y m+n−1−i = 0 · y m+n−i = 0, or m + n − 1 − i ≥ n, so xi y m+n−i−i = xi · 0 = 0, so x + y is nilpotent and N is an ideal, and by definintion a nil ideal. b) Let r + N be a nilpotent element of R/N , so there exists n ∈ Z+ such that rn ∈ N . But this means there exists m ∈ Z+ such that 0 = (rn )m = rnm , and thus r itself is a nilpotent element. c) In plainer terms: if S is a reduced ring and f : R → S is a ring homomorphism, then there exists a unique homomorphism f : R/N → S such that f = f ◦ q . Given this, the proof is straightforward, and we leave it to the reader. d) Suppose x is a nilpotent element of R, i.e., ∃n ∈ Z+ such that xn = 0. If ∩ is p a prime ideal, then since 0 = x · · · x ∈ p, we conclude x ∈ p: this shows N ⊂ p. Conversely, suppose x is not nilpotent. Then the set Sx := {xn | n ∈ N} satisfies 78 PETE L. CLARK (i) and (ii) of Proposition 117, so we may apply that result to get a prime ideal p which is disjoint from Sx , hence not containing x. Exercise 4.6: Prove Proposition 121c). An ideal I of a ring R is radical if for all x ∈ R, n ∈ Z+ , xn ∈ I implies x ∈ I . Exercise 4.7: a) Show that a prime ideal is radical. b) Exhibit a radical ideal which is not prime. c) Find all radical ideals in R = Z. d) Show that R is reduced iff (0) is a radical ideal. ∩ e) Let {Ii } be a set of radical ideals in a ring R. Show I = i Ii is a radical ideal. For any ideal I of R, we define the radical of I : r(I ) = {x ∈ R | ∃n ∈ Z+ xn ∈ I }. Proposition 122. Let R be a commutative ring and I , J ideals of R. a) r(I ) is the intersection of all prime ideals containing I , and is a radical ideal. b) (i) I ⊂ r(I ), (ii) r(r(I )) = r(I ); (iii) I ⊂ J =⇒ r(I ) ⊂ r(J ). c) r(IJ ) = r(I ∩ J ) = r(I ) ∩ r(J ). d) r(I + J ) = r(r(I ) + r(J )). e) r(I ) = R ⇐⇒ I = R. f ) For all n ∈ Z+ , r(I n ) = r(I ). Proof. First we make the following observation: under the canonical homomorphism q : R → R/I , r(I ) = q −1 (N (R/I )). By Proposition 113a), r(I ) is an ideal. a) Since N is the intersection of all prime ideals of R/I , r(I ) is the intersection of all prime ideals containing I , which is, by Exericse X.Xe), a radical ideal. b) (i) is immediate from the definition, and (ii) and (iii) follow from the characterization of r(I ) as the intersection of all radical ideals containing I . c) Since IJ ⊂ I ∩ J , r(IJ ) ⊂ r(I ∩ J ). If xn ∈ I ∩ J , then x2n = xn xn ∈ IJ , so x ∈ r(IJ ); therefore r(IJ ) = r(I ∩ J ). Since I ∩ J is a subset of both I and J , r(I ∩ J ) ⊂ r(I ) ∩ r(J ). Conversely, if x ∈ r(I ) ∩ r(J ), then there exist m and n such that xm ∈ I and xn ∈ J , so xmn ∈ I ∩ J and x ∈ r(I ∩ J ). d) Since I + J ⊂ r(I ) + r(J ), r(I + J ) ⊂ r(r(I ) + r(J )). A general element of r(I )+ r(J ) is of the form x + y , where xm ∈ I and y n ∈ J . Then (x + y )m+n ∈ I + J , so x + y ∈ r(I + J ). e) Evidently r(R) = R. Conversely, if r(I ) = R, then there exists n ∈ Z+ such that 1 = 1n ∈ I . f) By part a), r(I n ) is the intersection of all prime ideals p ⊃ I n . But by Corollary 119, a prime contains I n iff it contains I , so r(I n ) = r(I ). Remark: Proposition 122b) asserts that the mapping I → r(I ) is a closure operator on the lattice I (R) of ideals of R. Exercise 4.8: Let I be an ideal in the ring R. Show that r(I ) is the intersection of all prime ideals containing I . (Hint: reduce to the case I = 0.) An ideal p of a ring R is primary if every zero divisor of R/p is nilpotent. Equivalently, xy ∈ p, x ∈ p =⇒ y n ∈ p for some n ∈ Z+ . More on primary ideals in §X.X . / COMMUTATIVE ALGEBRA 79 We also define the Jacobson radical J (R) as the intersection of all maximal ideals of R. Evidently we have N ⊂ J (R). Proposition 123. Let R be a ring. An element x of R lies in the Jacobson radical J (R) iff 1 − xy ∈ R× for all y ∈ R. Proof. Suppose x lies in every maximal ideal of R. If there exists y such that 1 − xy is not a unit of R, then 1 − xy lies in some maximal ideal m, and then x ∈ m implies xy ∈ m and then 1 = (1 − xy ) + xy ∈ m, a contradiction. Conversely, suppose that there is a maximal ideal m of R which does not contain x. Then ⟨m, x⟩ = R, so 1 = m + xy for some m ∈ m and y ∈ R, and thus 1 − xy is not a unit. Remark: It is not yet clear why we have defined these two different notions of “radical.” Neither is it so easy to explain in advance, but nevertheless let us make a few remarks. First, the Jacobson radical plays a very important role in the theory of noncommutative rings, especially that of finite dimensional algebras over a field. (Indeed, a finite dimensional k -algebra is semisimple – i.e., a direct product of algebras without nontrivial two-sided ideals – iff its Jacobson radical is zero. In the special case of commutative algebras this comes down to the simpler result that a finite dimensional commutative k -algebra is reduced iff it is a product of fields.) Note that one important place in commutative algebra in which the Jacobson radical J (R) appears – albeit not by name, because of the necessity of putting the results in a fixed linear order – is in the statement of Nakayama’s Lemma. In general, the defining condition of nil(R) – i.e., as the intersection of all prime ideals of R – together with the fact that the radical of an arbitrary ideal I corresponds to the nilradical of R/I , makes the nilradical more widely useful in commutative algebra (or so it seems to the author of these notes). It is also important to consider when the nil and Jacobson radicals of a ring coincide. A ring R for which every homomorphic image S has nil(S ) = J (S ) is called a Jacobson ring; such rings will be studied in detail in §X.X . 4.3. Comaximal ideals. Recall (from Section X.X) that two ideals I and J in a commutative ring R are comaximal if I + J = R. Exercise 4.9: Determine all pairs of comaximal ideals in R = Z. Proposition 124. Let I and J be ideals in R. If r(I ) and r(J ) are comaximal, so are I and J . Proof. Apply Proposition 122d) and then Proposition 122e) to r(I ) + r(J ) = R: R = r(r(I ) + r(J )) = r(I + J ) = I + J. Recall that a set {Ii } of ideals of R is pairwise comaximal if for each i ̸= j , Ii + Ij = R. An immediate corollary of Proposition 124 is that if {Ii } are pairwise n comaximal and {ni } are any positive integers, then {Ii i } are pairwise comaximal. Lemma 125. Let J ,K1 , . . . , Kn be pairwise comaximal ideals in the ring R. Then ∩n K1 · · · Kn = i=1 Ki and J + K1 · · · Kn = R. 80 PETE L. CLARK Proof. We go by induction on n, the case of n = 1 being Lemma 26b). Suppose the theorem is true for any collection of n − 1 ideals K1 , . . . , Kn−1 and an ideal M such that the set {K1 , . . . , Kn−1 , M } is pairwise comaximal. First apply the ∩n−1 induction hypothesis with M = Kn to get that K ′ := K1 · · · Kn−1 = i=1 Ki and K ′ + Kn = R. Now part b) gives that K1 · · · Kn = K = K ′ Kn = K ′ ∩ Kn = n ∩ Ki . i=1 Now apply the induction hypothesis with M = J to get that K ′ = K1 · Kn−1 and J are comaximal, so so that there exist j ∈ J , k ′ ∈ K ′ such that j + k ′ = 1. Now let kn be any element of kn ; we have kn = jkn + k ′ kn ∈ J + K . Therefore J + K contains both J and Kn , but since J + Kn = R this implies J + K = R. Theorem 126. (Chinese Remainder Theorem, or “CRT”) Let R be a ring and I1 , . . . , In a finite set of pairwise comaximal ideals. Consider the natural map n ∏ Φ:R→ R/Ii , i=1 x → (x + Ii )n . Then Φ is surjective with kernel I1 · · · In , so that there is an i=1 induced isomorphism n ∏ ∼ Φ : R/(I1 · · · In ) → R/Ii . i=1 ∩n Proof. The map Φ is well-defined and has kernel i=1 Ii . Since the Ii ’s are pairwise ∩n comaximal, Lemma 125 gives i=1 Ii = I1 · · · In . So it remains to show that Φ is surjective. We prove this by induction on n, the case n = 1 being trivial. So we ∏n−1 may assume that the natural map Φ′ : R → R′ := i=1 R/Ii is surjective, with kernel I ′ := I1 · · · In−1 . Let (r′ , s) be any element of R′ × R/In . By assumption, there exists r ∈ R such that Φ′ (r + I ′ ) = r′ . Let s be any element of R mapping to s ∈ R/In . By Lemma 125, I ′ + In = R, so there exist x ∈ I ′ , y ∈ In such that s − r = x + y . Then Φ′ (r + x) = r′ , and r + x ≡ r + x + y ≡ s (mod In ), so Φ(r + x) = (r′ , s) and Φ is surjective. Remark: In the classical case R = Z, we can write Ii = (ni ) and then we are trying to prove – under the assumption that the ni ’s are coprime in pairs in the sense of elementary number theory – that the injective ring homomorphism Z/(n1 · · · nn ) → Z/n1 × · · · × Z/nn is an isomorphism. But both sides are finite rings of order n1 · · · nn , so since the map is an injection it must be an isomorphism! Nevertheless the usual proof of CRT in elementary number theory is much closer to the one we gave in the general case: in particular, it is constructive. Exercise 4.10: Let∏ be a ring and I1 , . . . , In any finite sequence of ideals. Consider R n the map Φ : R → i=1 R/Ii as in CRT. a) Show that Φ is surjective only if the {Ii } are pairwise comaximal. ∩n b) Show that Φ is injective iff i=1 Ii = (0). Exercise 4.11: a) Let G be a finite commutative group with exactly one element z of order 2. Show ∑ that x∈G x = z . COMMUTATIVE ALGEBRA 81 b) Let G be a finite commutative group which does not have exactly one element ∑ of order 2. Show that x∈G x = 0. c) Prove the following result of Gauss (a generalization of Wilson’s Theorem): let N ∈ Z+ , and put ∏ P (N ) = x. x∈(Z/N Z)× Then: P (N ) = ±1, and the minus sign holds iff N = 4 or is of the form pm or 2pm for an odd prime p and m ∈ Z+ . d) For a generalization to the case of (ZK /A)× , where A is an ideal in the ring ZK of integers of a number field K , see [Da09]. Can you extend Dalawat’s results to the function field case? Exercise 4.12: Let K be any field, and put R = K [t]. a) Let n1 , . . . , nk be a sequence of non-negative integers and {x1 , . . . , xk } a k element subset of K . For 1 ≤ i ≤ k , let ci0 , . . . , cini be a finite sequence of ni + 1 elements of k (not necessarily distinct). By applying the Chinese Remainder Theorem, show that there is a polynomial P (t) such that for 1 ≤ i ≤ k and 0 ≤ j ≤ ni we have P (j ) (xi ) = cij , where P (j ) (xi ) denotes the j th “formal” derivative of P evaluated at xi . Indeed, find all such polynomials; what can be said about the least degree of such a polynomial? b) Use the proof of the Chinese Remainder Theorem to give an explicit formula for such a polynomial P . Exercise 4.13: Let (M, ·) be a monoid and k a field. A character on M with values in k is a homomorphism of monoids from M to the multiplicative group k × of k . Each character lies in the k -vector space k M of all functions from M to k . a) (Dedekind) Show that any finite set of characters is k -linearly independent. b) Give an example of an infinite set of characters which is k -linearly dependent. Exercise 4.14: Show that for a ring R, TFAE: (i) R has finitely many maximal ideals.19 (ii) The quotient of R by its Jacobson radical J (R) is a finite product of fields. Exercise 4.15: Let R be a ring with the property that for every nonzero ideal I of R, R/I is finite.20 a) Show that every nonzero prime ideal of R is maximal. b) For a nonzero ideal I of R, define N (I ) = #R/I . Note that CRT implies that N is multiplicative in the sense that if I and J are comaximal then N (IJ ) = N (I )N (J ). In particular, show that this multiplicativity holds for any distinct nonzero prime ideals p and q. c) Show that for any n ∈ Z+ there are only finitely many ideals I with N (I ) ≤ n. We can therefore define a formal Dirichlet series ∑ ζR (s) = N (I )−s , I { 0} 19Such rings are typically called semilocal. I am not a fan of the terminology – it seems to either suggest that R has one half a maximal ideal (whatever that could mean) or two maximal ideals. But it is well entrenched, and I will not campaign to change it. 20We say that R has finite quotients. 82 PETE L. CLARK the Dedekind zeta function of R. d) Suppose that every nonzero ideal I of R factors uniquely into a product of prime ideals.21 Show that N is completely multiplicative in the sense that for all nonzero ideals I and J of R, N (IJ ) = N (I )N (J ). Deduce an Euler product factorization for ζR (s). 4.4. Local rings. All rings are commutative with unity. Proposition 127. For a ring R, TFAE: (i) There is exactly one maximal ideal m. (ii) The set R \ R× of nonunits forms a subgroup of (R, +). (iii) The set R \ R× is a maximal ideal. A ring satisfying these equivalent conditions is called a local ring. ∪ Proof. Since R× = R \ m m, the union extending over all maximal ideals of R, it follows that if there is only one maximal ideal m then m = R \ R× . This shows (i) =⇒ (iii) and certainly (iii) =⇒ (ii). Conversely, since the set of nonunits of a ring is a union of ideals, it is closed under multiplication by all elements of the ring. Thus it is itself an ideal iff it is an additive subgroup: (ii) =⇒ (iii). The implication (iii) implies (i) is very similar and left to the reader. Warning: In many older texts, a ring with a unique maximal ideal is called “quasilocal” and a local ring is a Noetherian quasi-local ring. This is not our convention. Local rings (especially Noetherian local rings) play a vital role in commutative algebra: the property of having a single maximal ideal simplifies many ideal-theoretic considerations, and many ring theoretic considerations can be reduced to the study of local rings (via a process called, logically enough, localization : see §X ). A field is certainly a local ring. The following simple result builds on this trivial observation to give some further examples of local rings: Proposition 128. Let I be an ideal in the ring R. a) If rad(I ) is maximal, then R/I is a local ring. b) In particular, if m is a maximal ideal and n ∈ Z+ then R/mn is a local ring. ∩ Proof. a) We know that rad(I ) = p⊃I p, so if rad(I ) = m is maximal it must be the only prime ideal containing I . Therefore, by correspondence R/I is a local ring. (In fact it is a ring with a unique prime ideal.) b) By Proposition 122f), r(mn ) = r(m) = m, so part a) applies. So, for instance, for any prime number p, Z/(pk ) is a local ring, whose maximal ideal is generated by p. It is easy to see (using, e.g. the Chinese Remainder Theorem) that conversely, if Z/(n) is a local ring then n is a prime power. Example: The ring Zp of p-adic integers is a local ring. For any field k , the ring k [[t]] of formal power series with coefficients in k is a local ring. Both of these rings are also PIDs. A ring which is a local PID is called a discrete valuation ring; these especially simple and important rings will be studied in detail later. 21We will see later that this is a class of rings called Dedekind rings which can be characterized in many other ways. COMMUTATIVE ALGEBRA 83 Exercise 4.16: Show that a local ring is connected, i.e., e2 = e =⇒ e ∈ {0, 1}. 5. Examples of rings 5.1. Rings of numbers. The most familiar examples of rings are probably rings of numbers, e.g. Z ⊂ Q ⊂ R ⊂ C. These are, respectively, the integers, the rational numbers, the real numbers and the complex numbers. For any positive integer N the ring integers modulo N , denoted Z/N Z. We assume that the reader has seen all these rings before. Historically, the concept of a ring as an abstract structure seems to have arisen as an attempt formalize common algebraic properties of number rings of various sorts. It is my understanding that the term “ring” comes from Hilbert’s Zahlring (“Zahl” means “number” in German). Indeed, various sorts of extension rings of C – most famously Hamilton’s quaternions H – have been referred to as systems of hypercomplex numbers. This terminology seems no longer to be widely used. The adjunction process gives rise to many rings and fields of numbers, as already √ seen in §2.2. For instance, for any nonsquare integer D, let D be a complex √ number whose square is D: then Z[ D] is an interesting ring. √ √ Exercise 5.1: Show that Z[ D] = {a + b D | a, b ∈ Z}. √ In√ particular, (Z[ D], +) ∼ (Z2 , +) as abelian groups, although not as rings, since = Z[ D] is an integral domain and Z2 , being a disconnected ring, has nontrivial idempotents. More generally, let K be any number field, i.e., a finite degree field extension of Q. Define ZK to be the set of elements x ∈ K which satisfy a monic polynomial with Z-coefficients. It turns out that ZK is a ring, the ring of algebraic integers in K. This is a very special case of the theory of integral closure, which we will study in great detail later on. Algebraic number theory proper begins with the observation that in general the rings ZK need not be UFDs but are otherwise as nice as possible from a commutative algebraic standpoint. That is, every ring ZK is a Dedekind domain, which among many other characterizations, means that every nonzero ideal factors into a product of prime ideals. That the rings ZK are Dedekind domains is an example of a normalization theorem, more specifically a very special case of the Krull-Akizuki Theorem, which we will meet later on. In particular, if one wanted to give an extremely laconic description of algebraic number theory – certainly incomplete, but rather accurate in spirit – one could say that it is the study of the Picard groups Pic ZK of rings of integers of number fields. Let Q be an algebraic closure of Q. (This is not a number field, being an infinite degree algebraic extension of Q.) We may define Z to be the set of all elements 84 PETE L. CLARK of Q which satsify a monic polynomial with integer coefficients: this is the ring of all algebraic integers. In particular, Z = lim ZK − → is the direct limit of all rings of integers in fixed number fields. This is a nice example of a ring which is natural enough but is “bigger” than the rings which were more classically studied. Exercise 5.2: Let Z be the set of all algebraic integers. a) Taking as given that for any fixed number field K , the algebraic integers in K form a subring of K , show that Z is a subring of Q. b) Show that Z is an integral domain which is not Noetherian. Hint: use the fact that the nth root of an algebraic integer is an algebraic integer to construct an infinite strictly ascending chain of principal ideals in Z. Theorem 129. Every finitely generated ideal in the ring Z is principal. Thus, if only Z were Noetherian, it would be a principal ideal domain! Later on we will prove a more general theorem, due to Kaplansky, in the context of limits of Dedekind domains with torsion Picard groups. 5.2. Rings of continuous functions. Let R be a ring, X a set, and consider the set RX of all functions f : X → R. We may endow RX with the structure of a ring by defining addition and multiplication “pointwise”, i.e., (f + g ) : x → f (x) + g (x), (f g ) : x → f (x)g (x). Exercise X.X: Show that this makes RX into a ring with additive identity the constant function 0 and multiplicative identity the constant function 1. However, this is not really a “new” example of a ring. Exercise 5.3: Show that RX is isomorphic as a ring, to ∏ x ∈X R. Later on we will see that this sort of construction is of some use in the special case that R = F , in which case we get an important class of Boolean rings. However, in general RX is quite a roomy ring. It contains many interesting subrings, some of which can be nicely consructed and analyzed using topological, geometric and analytic considerations. Suppose instead that we specialize to the following situation: R = R (the real numbers!), X is not just any old set but a topological space, and instead of the ring RX of all functions f : X → R we look instead at the subring C (X ) of continuous functions. Exercise 5.4: Show that for a topological space X , TFAE: (i) For every x, y ∈ X with x ̸= y , there exists f ∈ C (X ) with f (x) ̸= f (y ). (ii) For every x, y ∈ X with x ̸= y and every α, β ∈ R, there exists f ∈ C (X ) with f (x) = α, f (y ) = β . COMMUTATIVE ALGEBRA 85 (iii) For every finite subset S of X and any function g : S → R, there exists f ∈ C (X ) such that f |S = g . A space which satisfies these equivalent conditions is called C-separated.22 Recall the following chains of implications from general topology: Lemma 130. For any topological space, the following implications hold (and none of the arrows may be reversed)): a) X compact =⇒ X normal =⇒ X Tychonoff =⇒ X regular =⇒ X Hausdorff =⇒ X separated =⇒ X Kolmogorov. b) X locally compact =⇒ X Tychonoff. Exercise 5.5: a) Show that a Tychonoff space is C-separated. b) Show that a C-separated space is Hausdorff. c)* Show that a regular space need not be C-separated. (Suggestion: see [Ga71].) For a topological space X , a zero set is a set of the form f −1 (0) for some continuous function f : X → R. A cozero set is a complement of a zero set. The cozero sets in fact form a base for a topology on X , called (by us, at least) the Z-topology. Let us write XZ for X endowed with the Z -topology. Since every cozero set is an open set in the given topology on X , XZ is a coarser topology than the given topology on X : of course by this we allow the possibility that the two topologies coincide: i.e., every closed set is an intersection of zero sets of continuous R-valued functions. The following basic (but not so widely known) result gives a condition for this. Theorem 131. a) For a Hausdorff topological space X , the following are equivalent: (i) XZ = X : every closed set is an intersection of zero sets of continuous functions. (ii) X is Tychonoff, i.e., if Y is a closed subset of X and x ∈ X \ Y , then there exists a continuous function f : X → [0, 1] with f (x) = 0, f |Y ≡ 1. b) For any topological space X , the space XZ is completely regular, and is the finest completely regular topology on the underlying set of X which is coarser than X . Proof. Gilman-Jerison, p. 38. Let X be a topological space, and let x ∈ X be any point. Consider the set mx = {f ∈ C (X ) | f (x) = 0}. Evidently mx is an ideal of C (X ). But more is true. ∼ Proposition 132. Evaluation at x gives a canonical isomorphism C (X )/mx → R. In particular, mx is a maximal ideal of C (X ). Exercise 5.6: Prove Proposition 132. Thus x → mx gives a map of sets M : X → M (X ). Proposition 133. The map M : X → M (X ) is injective iff X is C-separated. Proof. This is left to the reader as a routine check on parsing the definitions. 22More standard terminology: “the continuous functions on X separate points”. 86 PETE L. CLARK Proposition 134. If X is quasi-compact, then M is surjective, i.e., every maximal ideal of C (X ) is of the form mx for at least one point x ∈ X . Proof. It suffices to show: let I be an ideal of C (X ) such that for no x ∈ X do we have I ⊂ mx . Then I = C (X ). By hypotehsis, for every x ∈ X there exists fx ∈ I such that fx (x) ̸= 0. Since fx is continuous, there exists an open neighborhood Ux of x such fx is nowhere vanishing on Ux . By quasi-compactness of X , there exists a finite set x1 , . . . , xN ∪N 2 2 such X = i=1 Uxi . Then the function f = fx1 + . . . + fxn is an element of m which 1 is strictly positive at every x ∈ X . But then f is also a continuous function on X , i.e., f ∈ C (X )× , so I = R. A compact space is quasi-compact and C-separated. Thus previous results yield: Theorem 135. If X is a compact space, then M : X → M (X ) is a bijection: every maximal ideal of the ring of continuous functions on X is of the form mx for a unique x ∈ X . In fact more is true. The perspective of functional analysis yields a natural topology on M (X ), the initial topology: namely, each f ∈ C (X ) induces a function Mf : M (X ) → R, namely Mf maps m to the image of f in C (X )/m = R. Now we endow M (X ) with the coarsest topology which makes each of the functions Mf continuous. Lemma 136. For a compact space X , the initial topology on M (X ) is Hausdorff. Proof. For distinct x, x′ ∈ X , consider the maximal ideals mx , mx′ . By C-separatedness, there exists f ∈ C (X ) with f (x) = 0, f (x′ ) ̸= 0. Thus choose disjoint neighborhoods V, V ′ of f (x), f (x′ ) ∈ R. The sets Uf,V = {x ∈ X | f (x) ∈ V }, Uf,V ′ = {x ∈ X | f (x) ∈ V ′ } are disjoint open neighborhoods of x and x′ . Theorem 137. For a compact space X , let M (X ) be the set of maximal ideals of C (X ) endowed with the initial topology. Then M : X → M (X ), x → mx is a homeomorphism. Proof. Step 1: We claim that M is continuous. But indeed, by the universal property of the initial topology, it is continuous iff for all f ∈ C (X ), the composite function x → mx → f (mx ) is continuous. But this is nothing else than the function x → f (x), i.e., the continuous function f ! So that was easy. Step 2: We now know that M is a continuous bijection from a compact space to a Hausdorff space. Therefore it is a closed map: if Y is closed in X , then Y is compact, so M(Y ) is compact, so M(Y ) is closed. Therefore M−1 is continuous and thus M is a homeomorphism. For any commutative ring R, we define MaxSpec(R) to be the maximal ideals and put a topology on it: for any ideal I of R, we define V (I ) = {m ∈ MaxSpec R | I ⊂ m} The sets V (I ) are the closed sets for a unique topology on MaxSpec R, the Zariski topology. Another way to say it is that the closed sets in the Zariski topology are precisely all sets obtained by intersecting sets of the form V (f ) = {m ∈ MaxSpec R | f ∈ m}. COMMUTATIVE ALGEBRA 87 To see this, note first that for any ideal I of R, ∩ V (I ) = V (f ) f ∈I and for any subset S of R, ∩ f ∈S V (f ) = ∩ V (f ). f ∈⟨S ⟩R Thus, from the perspective of the rest of these notes, it is natural to consider M (X ) = MaxSpec C (X ) as being endowed with the Zariski topology rather than the initial topology (note that the latter is defined only in the quasi-compact case). Proposition 138. Let X be any topological space. Then the map M : X → MaxSpec C (X ) is continuous when MaxSpec X is given the Zariski topology. Proof. As above, it is enough to show that for all f ∈ C (X ), the preimage M−1 (V (f )) is closed in X . Unpacking the definitions, we find M−1 (V (f )) = f −1 (0), thus the preimage is the zero set of the continuous function f , hence closed. Corollary 139. For a compact space X , the Zariski topology on M (X ) = MaxSpec C (X ) coincides with the initial topology. Proof. By Theorem 137, we may instead compare the Zariski topology on X – i.e., the topology obtained by pulling back the Zariski topology on MaxSpec C (X ) via M – with the given topology on X . But the proof of Proposition 138 shows that the Zariski topology on X is precisely the Z-topology, i.e., the one in which the closed subsets are the intersections of zero sets. But X is compact hence quasi-Tychonoff, so by Theorem 131 the Z-topology on X coincides with the given topology on X. Now let π : X → Y be a continuous map between compact spaces. There is an induced map C (π ) : C (Y ) → C (X ): given g : Y → R, we pullback by π to get g ◦ π : X → R. It is no problem to see that C (π ) is a homomorphism of rings. Now let mx ∈ MaxSpec C (X ) be a maximal ideal and consider its pullback C (π )∗ (mx ) to an ideal of C (Y ): we find C (π )∗ (mx ) = {g : Y → R | g (π (x)) = 0} = mπ(x) . Thus the pullback map carries maximal ideals to maximal ideals (recall this is certainly not true for all homomorphisms of rings!) and thus induces a map from X to Y which is indeed nothing else than the given map π . All in all we see that the functors C and MaxSpec give a duality between the categories of compact spaces and rings of continuous R-valued functions on compact spaces. In functional analysis this the first step in an important circle of ideas leading up to Gelfand duality for commutative Banach algebras. What about the case of noncompact spaces X ? Example: Let X be an infinite discrete space, so C (X ) = RX is the ring of all functions from X to R. Thus X is a noncompact Tychonoff space. So it follows 88 PETE L. CLARK from our work so far that M gives a continuous injection from M to the quasicompact space MaxSpec C (X ). In fact M is an embedding: for any subset Y ⊂ X , let IY be the ideal of functions vanishing identically on Y . Then the restriction of the closed subset V (I ) of MaxSpec C (X ) to M(X ) is precisely M(Y ), so M(Y ) is closed in MaxSpec C (X ). Thus M(X ) is discrete as a subspace of MaxSpec C (X ), and this implies that M is not surjective. Theorem 140. For any topological space X , MaxSpec C (X ) endowed with the Zariski topology is compact. Theorem 141. Let X be any topological space, let M : X → MaxSpec C (X ), and let XT = M(X ), viewed as a subspace of MaxSpec C (X ). a) XT is a Tychonoff space. b) The map M : X → XT is the Tychonoff completion of X : i.e., it is universal for continuous maps from X to a Tychonoff space. c) The induced map C (M) : C (XT ) → C (X ) is an isomorphism of rings. Proof. See [GJ76, §3.9]. Thus the ring of continuous functions on an arbitrary space X “sees” precisely its Tychonoff completion XT . Henceforth we restrict to Tychonoff spaces. Theorem 142. Let X be a Tychonoff space. Then the map M : X → C (X ) is nothing else than the Stone-Cech compactification. Proof. See [GJ76, §7.10]. Exercise 5.7: a) Show that C (X ) is an R-subalgebra of RX . b) Show that C (X ) is reduced: it contains no nonzero nilpotent elements. Exercise 5.8: Show that C (X ) is connected in the algebraic sense – i.e., there are no idempotents other than 0 and 1 – iff the topological space X is connected. Exercise 5.9: Show that there is an antitone Galois connection between 2X and the set of ideals of C (X ), as follows: S ⊂ X → IS = {f ∈ C (X ) | f |S ≡0 } and I → VI = {x ∈ X | ∀f ∈ I, f (x) = 0}. Exercise 5.10: Let X be a compact space. a) Let p be a prime ideal of C (X ). Show that V (p) consists of a single point. b) Deduce that a prime ideal p of C (X ) is closed in the sense of the Galois connection – i.e., p = IVp iff p is maximal. c) Deduce that each prime ideal p of C (X ) is contained in a unique maximal ideal. Exercise 5.11: Let X = [0, 1] with the standard Euclidean topology. Let r0 be (x the ideal of all functions f ∈ C (X ) such that for all k ∈ N, limx→0+ fxk ) = 0. Equivalently r0 is the ideal of all functions which are infinitely differentiable at 0 and have identically zero Taylor series at zero. a) Show that r0 is a radical ideal but not a prime ideal. b) Show that the only maximal ideal containing r0 is m0 , the set of all functions vanishing at 0. c) Deduce that there exist ideals of C (X ) which are prime but not maximal. COMMUTATIVE ALGEBRA 89 Exercise 5.12: Let X be a C-separated topological space. a) Let S ⊂ X with #S > 1. Show that IS is not maximal. b) Suppose X is Tychonoff and S, T ⊂ X . Show that IS ⊂ IT ⇐⇒ T ⊂ S . c) In particular if X is Tychonoff, then for closed subsets S and T of X , IS = IT ⇐⇒ S = T . Exercise 5.13: Let φ : X → Y be a continuous function between topological spaces. a) Show that φ induces a ring homomorphism C (φ) : C (Y ) → C (X ) by g ∈ C (Y ) → φ∗ g = g ◦ φ. b) Suppose Y is normal, X is a closed subspace of X and φ : X → Y is the inclusion map. Show that C (φ) is surjective. Exercise 5.14: Let X be a normal topological space. Show that the closure operator on subsets of X given by the Galois connection coincides with the topological closure operator on X . 1 Exercise 5.15: Let X be the subspace { n }n∈Z+ ∪{0} of R, and let m be the maximal ideal of all functions vanishing at 0. Fill in the details of the following outline of a proof that m is not finitely generated.23 Assume otherwise: m = ⟨a1 , . . . , an ⟩. g2 ( Then for every element g ∈ m, limx→0 |a1 (x)|+...x)|an (x)| = 0. (In particular, there + exists δ > 0 such that the denominator is strictly positive on (0, δ ).) Now choose g ∈ m so as to get a contradiction. Exercise 5.16: Let X be a normal space, and let x ∈ X . a) Show that the following are equivalent: (i) The ideal Ix is finitely generated. (ii) The ideal Ix is principal. (iii) The point x is isolated in X (i.e., {x} is open). b) Suppose X is compact. Show that the following are equivalent: (i) C (X ) is a Noetherian ring. (ii) C (X ) is finite-dimensional as an R-vector space. (iii) X is finite. Exercise 5.17: Show that if we worked throughout with rings C (X, C) of continuous C-valued functions, then all of the above results continue to hold. Exercise 5.18: Suppose that we looked at rings of continuous functions from a topological space X to Qp . To what extent to the results of the section continue to hold? Exercise 5.19: Let X be a compact smooth manifold and consider the ring C∞ (X ) of smooth functions f : X → R. a)* Show that for x ∈ X , the ideal mx of all functions vanishing at x is maximal and finitely generated. b) Note that the phenomenon of part a) is in contrast to the case of maximal ideals in the ring C ([0, 1]), say. However, I believe that with this sole exception, all of the 23Or, if you like, give your own proof that m is not finitely generated. 90 PETE L. CLARK results of this section hold for the rings C∞ (X ) just as for the rings C (X ). Try it and see. 5.3. Rings of holomorphic functions. As we saw in the previous section, one of the characteristic properties of the ring of continuous functions on a normal space (or even smooth functions on a manifold) is that it is typically very far from being an integral domain. A remedy for this is to consider more “rigid” collections of functions. Let U be an open subset of the complex plane C, and let Hol(U ) be the set of holomorphic functions f : U → C. (Recall that a holomorphic function on U is one for which the complex derivative f ′ (z ) exists for each z ∈ U . Equivalently, for each z ∈ U f admits a power series development with positive radius of convergence.) It is immediate that Hol(U ) is a subring of the ring CU of all C-valued functions on U . Proposition 143. For a nonempty open subset U of C, the following are equivalent: (i) U is connected. (ii) Hol(U ) is a domain. Proof. (i) =⇒ (ii): For any f ∈ C (U, C) let Z (f ) = {z ∈ U | f (z ) = 0} be the zero set of f . Since f is continuous, Z (f ) is a closed subset of U . If f is moreover holomorphic, then Z (f ) has no accumulation point in U , i.e., f ̸= 0 =⇒ Z (f ) is discrete – 24 in particular Z (f ) is countable. Moreover, for any f, g ∈ C (U, C) we have Z (f g ) = Z (f ) ∪ Z (g ), so if f, g ∈ Hol(U )• then Z (f g ) is at most countable, whereas U is uncountable, so f g ̸= 0. (ii) =⇒ (i): we argue by contrapositive. If U is not connected, it is of the form V1 ∪ V2 where V1 and V2 are disjoint open subsets. Let χi be the characteristic function of Vi for i = 1, 2. Then each χi is locally constant on U – hence holomorphic, and nonzero, but χ1 χ2 = 0. Recall that in complex function theory it is common to call a nonempty open subset U of C a domain. In this language, Proposition 143 simply asserts that U is a domain iff Hol(U ) is a domain. Henceforth we assume that U is a domain. For every z ∈ U there is a function ordz : Hol(U )• → N, the order of vanishing ∑∞ of f at z. Precisely, we exand f into a power series at z : f (ζ ) = n=0 an (ζ − z )n and put ordz (f ) to the least n for which an ̸= 0. Compiling all these together we may associate to each f ∈ Hol(U )• its total order Ord(f ) : U → N given by Ord(f )(z ) = ordz (f ). Consider the set NU of all functions from U to N. For O ∈ NU , we define the support of O to be the set of z ∈ U such that O(z ) > 0. Recall that a meromorphic function on U is a function which is holomorphic on U except for isolated finite order singularities. More precisely, a meromorphic function is a function which is holomorphic on U \ Z for some discrete closed subset Z of U 24Recall that this is proved by considering the Taylor series development of f about any accumulation point. COMMUTATIVE ALGEBRA 91 and such that for all z0 ∈ Z , there exists n ∈ Z+ such that (z − z0 )n f (z ) extends to a holomorphic function on a neighborhood of z . If the least n as above is positive, we say that f has a pole at z0 , and we employ the convention that f (z0 ) = ∞. Let Mer(U ) be the set of all meromorphic functions on U ; it is a ring under pointwise addition and multiplication, under the conventions that for all z ∈ C, z + ∞ = ∞ + ∞ = z · ∞ = ∞ · ∞ = ∞. Theorem 144. Let U be a domain in the complex plane. a) (Weierstrass) For each O ∈ NU with closed, discrete support, there exists f ∈ Hol(U )• with Ord(f ) = O. b) (Weierstrass + Mittag-Leffler) Let Z ⊂ U be a closed subset without limit points. To each z ∈ Z we associate a natural number nz and for all 0 ≤ k ≤ nz , a complex number wz,k . Then there exists f ∈ Hol(U ) such that for all z ∈ Z and 0 ≤ k ≤ nz , f (k) (z ) = k !wz,k . Proof. Part a) is of the two main results in Weierstrass’ Factorization Theory: see e.g. [Ru87, Thm. 15.11]. Part b) is proved by combining part a) with MittagLeffler’s famous result on the existence of meromorphic functions with prescribed principal parts: see e.g. [Ru87, Thm. 15.13]. Corollary 145. The ring Mer(U ) of meromorphic functions on U is a field, and indeed is the field of fractions of Hol(U ). Exercise 5.20: Prove Proposition 145. Exercise 5.21: Fix z0 ∈ U . For f ∈ Mer(U ), choose n ∈ N such that (z − z0 )n f is holomorphic at z0 , and put ordz0 (f ) = ordz0 ((z − z0 )n f ) − n. a) Show that this gives a well-defined function ordz0 : Mer(U )• → Z (i.e., independent of the choice of n in the definition). b) Show that for all f, g ∈ Mer(U )× , ordz0 (f g ) = ordz0 (f ) + ordz0 (g ). c) We formally extend ordz0 to a function from Mer(U ) to Z ∪ {∞} by setting ordz0 (0) = ∞. Show that, under the convention that ∞ + n = ∞ + ∞ = ∞, we have for all f, g ∈ Mer(U ) that ordz0 (f + g ) ≥ min ordz0 (f ), ordz0 (g ). d) Show that if ordz0 (f ) ̸= ordz0 (g ) then ordz0 (f + g ) = min ordz0 f, ordz0 (g ). Similarly we may extend Ord to a function from Mer(U )• to ZU . Lemma 146. For f, g ∈ Hol(U )• , TFAE: (i) Ord(f ) = Ord(g ). (ii) f = ug for u ∈ Hol(U )× . (iii) (f ) = (g ). Proof. (ii) ⇐⇒ (iii) for elements of any integral domain. (ii) =⇒ (i) is easy and left to the reader. (i) =⇒ (ii): The meromorphic function f has identically zero order, hence is g nowhere vanishing and is thus a unit u in Hol(U ). Theorem 147. (Helmer [Hel40]) For a domain U in the complex plane, every finitely generated ideal of Hol(U ) is principal. More precisely, for any f1 , . . . , fn ∈ Hol(U )• , there exists f ∈ Hol(U ) such that Ord(f ) = mini Ord(fi ), unique up to associates, and then ⟨f1 , . . . , fn ⟩ = ⟨f ⟩. 92 PETE L. CLARK Proof. Step 1: Suppose that f1 , f2 ∈ Hol(U )• do not simultaneously vanish at any point of U . We claim that ⟨f1 , f2 ⟩ = Hol(R). proof of claim Let Z be the zero set of f1 , so that for all z ∈ Z , f2 (z ) ̸= 0. By Theorem 144b) there exists g2 ∈ Hol(U ) such that for all z ∈ Z , ordz (1 − g2 f2 ) ≥ g2 ordz (f1 ). Thus Ord(1 − g2 f2 ) ≥ Ord(f1 ), so that g1 := 1−f1 f2 ∈ Hol(U ) and thus f1 g1 + f2 g2 = 1. Step 2: Now let f1 , f2 ∈ Hol(U )• be arbitrary. By Theorem 144a), there exists f ∈ Hol(U ) with Ord(f ) = min Ord(f1 ), Ord(f2 ). For i = 1, 2, put gi = fi . Then g1 f and g2 are holomorphic and without a common zero, so by Step 1 ⟨g1 , g2 ⟩ = Hol(U ). Multiplying through by f gives ⟨f1 , f2 ⟩ = ⟨f ⟩. Step 3: An easy induction argument shows that in a ring R in which every ideal of the form ⟨x1 , x2 ⟩ is principal, every nitely generated ideal is principal. By Step 2, this applies in particular to Hol(U ). Moreover, it is easy to see that if the ideal ⟨f1 , . . . , fn ⟩ is generated by any single element f , then we must have Ord f = min Ord fi . Exercise 5.22: Explain carefully why in Step 1 of the above proof, Theorem 144b) implies the existence of g2 . Of course the most familiar class of domains in which every finitely generated ideal is principal are those domains in which every ideal is principal: PIDs! But as the reader has probably already suspected, Hol(U ) is not a PID. One way to see this is to show that Hol(U ) is not even a UFD. Remarkably, this is an immediate consequence of the Weierstrass Factorization Theory, which succeeds in decomposing every holomorphic function into a product of prime elements! The catch is that most holomorphic functions require infinite products, a phenomenon which is not countenanced in the algebraic theory of factorization. Exercise 5.23: Let f ∈ Hol(U )• . a) Show that f is an irreducible element of Hol(U ) – i.e., if f = g1 g2 then exactly one of g1 , g2 is a unit – iff it has exactly one simple zero. b) Suppose f is irreducible. Show that Hol(U )/(f ) = C. In particular, (f ) is a prime ideal. c) Show that f admits a (finite!) factorization into irreducible elements iff it has only finitely many zeros. Conclude that Hol(U ) is not a UFD. Exercise 5.24: Extract from the previous exercise an explicit ideal of Hol(C) which is not finitely generated. Exercise 5.25*: Show that all of the results of this section extend to the ring of holomorphic functions on a noncompact Riemann surface. Exercise 5.26*: Investigate the extent to which the results of this section continue to hold for Stein manifolds. (Step 1: learn the definition of a Stein manifold!) 5.4. Polynomial rings. COMMUTATIVE ALGEBRA 93 Let R be a ring (possibly non-commutative, but – as ever – with identity). Then R[t] denotes the ring of univariate polynomials with R-coefficients. We assume the reader knows what this means in at least an informal sense: an element of R will be an expression of the form an tn + . . . + a1 t + a0 , where n is some non-negative integer and an , . . . , a0 are in R. The degree of a polynomial is the supremum over all numbers n such that an ̸= 0. We say “supremum” rather than “maximum” as an attempt to justify the convention that the degree of the 0 polynomial should be −∞ (for that is the supremum of the empty set). A polynomial of degree 0 is called constant, and we can view R as a subset of R[t] by mapping a ∈ R to the constant polynomial a. As an abelian group, R[t] is canonically ⊕∞ ∑ n isomorphic to n=0 R, the isomorphism being given by n an t → (a0 , a1 , . . .). (The key point here is that on both sides we have an = 0 for all sufficiently large n.) Multiplication of polynomials is obtained by applying the relations t0 = 1, ti+j = ti tj at = ta for all a ∈ R, and distributivity, i.e., ∑ (an tn + . . . + an t1 + a0 ) · (bm tm + . . . + b1 t + b0 ) = ai bj ti+j . 0≤i≤n, 0≤j ≤m For any P ∈ R[t], the identity 1 ∈ R has the property 1 · P = P · 1 = 1. Unfortunately there are some minor annoyances of rigor in the previous description. The first one – which a sufficiently experienced reader will immediately either dismiss as silly or know how to correct – is that it is not set-theoretically correct : technically speaking, we need to say what R[t] is as a set and this involves saying what t “really is.” It is common in abstract algebra to refer to t is an indeterminate, a practice which is remarkably useful despite being formally meaningless: essentially it means “Don’t worry about what t is; it can be anything which is not an element of R. All we need to know about t is encapsulated in the multiplication rules at = ta, t0 = 1, ti tj = ti+j .” In other words, t is what in the uncomplicated days of high school algebra was referred to as a variable. If someone insists that R[t] be some particular set – a rather unenlighened attitude that we will further combat later on – then the solution has already been ⊕∞ given: we can take R[t] = n=0 R. (It is fair to assume that we already know what direct sums of abelian groups “really are”, but in the next section we will give a particular construction which is in fact rather useful.) This disposes of the set-theoretic objections. Not to be laughed away completely is the following point: we said R[t] was a ring, but how do we know this? We did explain the group structure, defined a multiplication operation, and identified a multiplicative identity. It remains to verify the distributivity of multiplication over addition (special cases of which motivated our definition of multiplication, but nevertheless needs to be checked in general) and also the associativity of multiplication. Neither of these properties are at all difficult to verify. In fact: Exercise 5.27: a) Show that R[t] is a ring. b) Show that R[t] is commutative iff R is commutative. 94 PETE L. CLARK Let us now attempt a “conceptual proof” of the associativity of polynomial multiplication. For this we shall assume that R is commutative – this is the only case we will be exploring further anyway. Then we can, as the P (t) notation suggests, view an element of R[t] as a function from R to R. Namely, we just plug in values: a ∈ R → P (a) ∈ R. To be clear about things, let us denote this associated function from R to R by P . As we saw above, the set of all functions RR from R to R forms a commutative ring under pointwise addition and multiplication: (f + g )(a) := f (a) + g (a), (f g )(a) := f (a) · g (a). In particular, it really is obvious that the multiplication of functions is associative. Let P be the subset of RR of functions of the form P for some P ∈ R[t]. More concretely, we are mapping the constant elements of R[t] to constant functions and mapping t to the identity function. This makes it clear that P is a subring of RR : in fact it is the subring of RR generated by the constant functions and the identity function. So why don’t we just define R[t] to be P , i.e., identify a polynomial with its associated function? The problem is that the map R[t] → P need not be an injection. Indeed, if R is finite (but not the zero ring), P is a subring of the finite ring RR so is obviously finite, whereas R[t] is just as obviously infinite. If R is a domain this turns out to be the only restriction. Proposition 148. Let R be an integral domain. a) Suppose that R is infinite. Then the canonical mapping R[t] → P is a bijection. b) Suppose that R is finite, say of order q , and is therefore a field. Then the kernel of the canonical mapping R[t] → P is the principal ideal generated by tq − t. We leave the proof as a (nontrivial) exercise for the interested reader. Exercise 5.28: Exhibit an infinite commutative ring R for which the map R[t] → P is not injective. (Suggestion: find an infinite ring all of whose elements x satisfy x2 = x.) Exercise 5.29: Show that the map R[t] → P is a homomorphism of rings. So if we restrict to infinite integral domains, the map R[t] → P is an isomorphism of rings. Thus we see, after the fact, that we could have defined the ring structure in terms of pointwise multiplication. 5.5. Semigroup algebras. A semigroup M is a set equipped with a single binary operation ·, which is required (only!) to be associative. A monoid is a semigroup with a two-sided identity. Exercise 5.30: Show that a semigroup has at most one two-sided identity, so it is unambiguous to speak of “the” identity element in a monoid. We will denote it by e (so as not to favor either addditive or multiplicative notation). COMMUTATIVE ALGEBRA 95 Example: Let (R, +, ·) be an algebra. Then (R, ·) is a semigroup. If R is a ring (i.e., has an identity 1) then (R, ·) is a monoid, with identity element 1. Example: Any group is a monoid. In fact a group is precisely a monoid in which each element has a two-sided inverse. Example: The structure (N, +) of natural numbers under addition is a monoid; the identity element is 0. Example: The structure (Z+ , ·) of positive integers under multiplication is a monoid; the identity element is 1. Let M and N be two semigroups. Then the Cartesian product M × N becomes a semigroup in an obvious way: (m1 , n1 ) · (m2 , n2 ) := (m1 · m2 , n1 · n2 ). If M and N are monoids with identity elements eM and eN , then M × N is a monoid, with identity element (eM , eN ). Exactly the same discussion holds⊕ any finite for n set M1 , . . . , MN of semigroups: we can form the direct sum M = i=1 Mi , i.e., the Cartesian product of sets with componentwise operations; if all the Mi ’s are monoids, so is M . If we instead have an infinite family {Mi }i∈I of semigroups indexed by a set I , we ∏ can define a semigroup structure on the Cartesian product i∈I Mi in the obvious way, and if each Mi is a monoid with identity ei , then the product semigroup is a monoid with identity (ei )i∈I . If each Mi is a monoid, we can also define the direct ⊕ ∏ sum i∈I Mi , which is the subset of the direct product i∈I Mi consisting of all I -tuples (mi ∈ Mi )i∈I such that mi = ei for all but finitely many i. Then we have ⊕ ∏ that i∈I Mi is a submonoid of the direct product monoid i∈I Mi . If M and N are semigroups, then a map f : M → N is a homomorphism of semigroups if f (m1 · m2 ) = f (m1 ) · f (m2 ) for all m1 , m2 ∈ M . If M and N are monoids, a homomorphism of monoids is a homomorphism of semigroups such that moreover f (eM ) = eN . A homomorphism f : M → N of semigroups (resp. of monoids) is an isomorphism iff there exists a homomorphism of semigroups (resp. monoids) g : N → M such that g ◦ f = IdM , f ◦ g = IdN . Exercise 5.31: a) Exhibit monoids M and N and a homomorphism of semigroups f : M → N which is not a homomorphism of monoids. b) Show that a homomorphism of semigroups f : M → N is an isomorphism iff it is bijective. Same for monoids. Exercise 5.32: Show that the monoid (Z+ , ·) of positive integers under multipli⊕∞ cation is isomorphic to i=1 (N, +), i.e., the direct sum of infinitely many copies of the natural numbers under addition. (Hint: a more natural indexing set for the direct sum is the set of all prime numbers.) Now let R be an algebra and M be a semigroup. We suppose first that M is finite. Denote by R[M ] the set of all functions f : M → R. 96 PETE L. CLARK As we saw, using the operations of pointwise addition and multiplication endow this set with the structure of an associative algebra (which has an identity iff M does). We are going to keep the pointwise addition but take a different binary operation ∗ : R[M ] × R[M ] → R[M ]. Namely, for f, g ∈ R[M ], we define the convolution product f ∗ g as follows: ∑ (f ∗ g )(m) := f (a)g (b). (a,b)∈M 2 | ab=m In other words, the sum extends over all ordered pairs (a, b) of elements of M whose product (in M , of course), is m. Proposition 149. Let R be an associative algebra and M a finite semigroup. The structure (R[M ], +, ∗) whose underlying set is the set of all functions from M to R, and endowed with the binary operations of pointwise additition and convolution product, is an associative algebra. If R is a ring and M is a monoid with identity e, then R[M ] is a ring with multiplicative identity the function I which takes eM to 1R and every other element of M to 0R . Proof. First, suppose that R is a ring and M is a monoid, then for any f ∈ R[M ] and m ∈ M , we have ∑ (f ∗I )(m) = f (a)I (b) = f (m)I (1) = f (m) = I (1)f (m) = . . . = (I ∗f )(m). (a,b)∈M 2 | ab=m We still need to check the associativity of the convolution product and the distributivity of convolution over addition. We leave the latter to the reader but check the former: if f, g, h ∈ R[M ], then ∑ ∑∑ ((f ∗ g ) ∗ h)(m) = (f ∗ g )(x)h(c) = f (a)g (b)h(c) xc=m = = ∑∑ ay =m bc=y ∑ xc=m ab=x f (a)g (b)h(c) abc=m f (a)g (b)h(c) = ∑ f (a)(g ∗ h)(y ) = (f ∗ (g ∗ h))(m). ay =m A special case of this construction which is important in the representation theory of finite groups is the ring k [G], where k is a field and G is a finite group. Now suppose that M is an infinite semigroup. Unless we have some sort of extra structure on R which allows us to deal with convergence of sums – and, in this level of generality, we do not – the above definition of the convolution product f ∗ g is problematic because the sum might be infinite. For instance, if M = G is any group, then our previous definition of (f ∗ g )(m) would come out to be ∑ −1 m), which is, if G is infinite, an infinite sum. x∈G f (x)g (x Our task therefore is to modify the construction of the convolution product so as to give a meaningful answer when the semigroup M is infinite, but in such a way that agrees with the previous definition for finite M . COMMUTATIVE ALGEBRA 97 Taking our cue from the infinite direct sum, we restrict our domain: define R[M ] to be subset of all functions f : M → R such that f (m) = 0 except for finitely many m (or, for short, finitely nonzero functions). Restricting to such functions, ∑ (f ∗ g )(m) := f (a)g (b) ab=m makes sense: although the sum is apparently infinite, all but finitely terms are zero. Proposition 150. Let R be an associative algebra and M a semigroup. The structure (R[M ], +, ∗) whose underlying set is the set of all finitely nonzero functions from M to R, and endowed with the binary operations of pointwise additition and convolution product, is an associative algebra. If R is a ring and M is a monoid with identity element e, then R[M ] is a ring with multiplicative identity the function I which takes eM to 1R and every other element of M to 0R . Exercise 5.33: Prove Proposition 150. More precisely, verify that the proof of Proposition 149 goes through completely unchanged. Note that as an abelian group, R[M ] is naturally isomorphic to the direct sum ⊕ m∈M R, i.e., of copies of R indexed by M . One can therefore equally well view ∑ an element R[M ] as a formal finite expressions of the form m∈M am m, where am ∈ R and all but finitely many are 0. Written in this form, there is a natural way to define the product ( )( ) ∑ ∑ am m bm m m∈M m∈M of two elements f and g of R[M ]: namely we apply distributivity, use the multiplication law in R to multiply the am ’s and the bm ’s, use the operation in M to multiply the elements of M , and then finally use the addition law in R to rewrite ∑ the expression in the form m cm m. But a moment’s thought shows that cm is nothing else than (f ∗ g )(m). On the one hand, this makes the convolution product look very natural. Conversely, it makes clear: The polynomial ring R[t] is canonically isomorphic to the monoid∑ ring R[N]. Indeed, the explict isomorphism is given by sending a polynomial n an tn to the function n → an . This gives a new proof of the associativity of the product in the polynomial ring R[t]. We leave it to the reader to decide whether this proof is any easier than direct verification.. Rather the merit is that this associativity computation has been done once and for all in a very general context. As an aside, let me point out something very curious: in searching for a slick proof of associativity of multiplication in the polynomial ring R[t], we attempted to show that the multiplication was just multiplication of the associated functions f : R → R. As we saw, this works in many but not all cases (because the homomorphism from R[t] to the ring of functions RR is not always surjective). Then we observed that the associativity of multiplication of polynomials is a special case of associativity of the product in a semigroup ring. What is strange is that the 98 PETE L. CLARK elements of this semigroup ring R[N] are themselves defined as functions, but functions from N to R, and the product is not the most obvious (pointwise) one – which has nothing to do with the semigroup structure on N – but rather a “funny” convolution product. Suitably mathematically urbane readers will, upon seeing a homomorphism from an abelian group (here R[t]) to another abelian group (here P ) which converts a “convolution product” on the first group to a “pointwise product” on the second group, be tempted to view the mapping R[t] → P as some sort of Fourier transform. If I understood this phenomenon more completely myself, I might be more willing to digress to explain it (but I don’t, so I won’t). The semigroup algebra construction can be used to define several generalizations of the polynomial ring R[t]. Exercise 5.34: For any ring R, identify the monoid ring R[Z] with the ring R[t, t−1 ] of Laurent polynomials. ⊕ First, let T = {ti } be a set. Let F A(T ) := i∈T (N, +) be the direct sum of a number of copies of (N, +) indexed by T . Let R be a ring, and consider the monoid ring R[F A(T )]. Let us write the composition law in F A(T ) multiplicatively; moreover, viewing an arbitrary element I of F A(T ) as a finitely nonzero function from T ∏ to N, we use the notation tI for t∈T tI (t) . Then an arbitrary element of R[F A(T )] ∑n is a finite sum of the form k=1 rk tIk , where I1 , . . . , Ik are elements of F A(t). This representation of the elements should make clear that we can view R[F A(T )] as a polynomial ring in the indeterminates t ∈ T : we use the alternate notation R[{ti }]. Let us go back to the monoid ring R[N], whose elements are finitely nonzero functions f : R → N. Notice that in this case the precaution of restricting finitely nonzero functions is not necessary: the monoid (N, +), although infinite, has the property that for any m ∈ N, the set of all x, y ∈ N such that x + y = m is finite (indeed, of cardinality m + 1). Let us call an arbitrary monoid M divisor-finite if for each m in M , the set {(x, y ) ∈ M 2 | xy = m} is finite. Exercise 5.35: a) For any set T , the monoid F A(T ) = finite. b) A group is divisor-finite iff it is finite. ⊕ t∈T (N, +) is divisor- For a divisor-finite monoid M , and any ring R, we may define the big monoid ring R[[M ]] to be the collection of all functions M → R, with pointwise addition and convolution product. For example, if M = (N, +), then writing M multiplicatively with n ∈ N → tn for some formal generator t, an element of the ring R[[M ]] is an infinite formal sum ∑ n n∈N rn t . Such sums are added coordinatewise and multiplied by distributivity: ( ∑ n∈N rn tn )( ∑ n∈N sn tn ) = r0 s0 + (r0 s1 + r1 s0 )t + . . . + ( n ∑ rk sn−k )tn + . . . . k=0 This ring is denoted by R[[t]] and called the formal power series ring over R. COMMUTATIVE ALGEBRA 99 Exercise 5.36: Using Exercise 5.35, define, for any set T = {ti } and any ring R, a formal power series ring R[[{ti }]]. Here is yet another variation on the construction: suppose M is a commutative, cancellative divisor-finite monoid endowed with a total order relation ≤. (Example: (N, +) or F A(T ) for any T .) There is then a group completion G(M ) together with an injective homomorphism of monoids M → G(M ). If M is finite and cancellative, it is already a group. If M is infinite, then so is G(M ), so it cannot be divisor-finite. Nevertheless, the ordering ≤ extends uniquely to an ordering on G(M ), and we can define a ring R((G(M )) whose elements are the functions from f : G(M ) → R such that {x ∈ G(M ) | x < 0, f (x) ̸= 0} is finite, i.e., f is finitely nonzero on the negative values of G(M ). Exercise 5.37: a) Show that under the above hypotheses, the convolution product on R((G(M )) is well-defined, and endows R((G(M )) with the structure of a ring. b) When M = (N, +), identify R((M )) as R((t)), the ring of formal finite-tailed Laurent series with coefficients in R. Give a multi-variable analogue of this by taking M = F A(T ) for arbitrary T . Exercise 5.38: Let R be a possibly non-commutative ring. Give a rigorous definition of the ring R⟨t1 , t2 ⟩ of “noncommutative polynomials” – each ti commutes with each element of R, but t1 and t2 do not commute – as an example of a small monoid ring R[M ] for a suitable monoid M . Same question but with an arbitrary set T = {ti } of noncommuting indeterminates. The universal property of semigroup rings: Fix a commutative ring R. Let B be a commutative R-algebra and M a commutative monoid. Let f : R[M ] → B be an R-algebra homomorphism. Consider f restricted to M ; it is a homomorphism of monoids M → (B, ·). Thus we have defined a mapping HomR-alg (R[M ], B ) → HomMonoid (M, (B, ·)). Interestingly, this map has an inverse. If g : M → B is any homomorphism satisfying g (0) = 0, g (m1 + m2 ) = g∑ 1 ) + g (m2 ), then g extends to a unique R-algebra (m ∑ homomorphism R[M ] → B : m∈M rm m → m rm g (m). The uniqueness of the extension is immediate, and that the extended map is indeed an R-algebra homomorphism can be checked directly (please do so). In more categorical language, this canonical bijection shows that the functor M → R[M ] is the left adjoint to the forgetful functor (S, +, ·) → (S, ·) from R-algebras to commutative monoids. Yet further terminology would express this by saying that R[M ] is a “free object” of a certain type. Theorem 151. (Universal property of polynomial rings) Let T = {ti } be a set of indeterminates. Let R be a commutative ring, and S an R-algebra. Then each map of sets T → S extends to a unique R-algebra homomorphism R[T ] → S . Proof: By the previous result, each monoid map from the free commutative monoid ⊕ t∈T Z to S extends to a unique R-algebra homomorhpism. So what is needed is the fact that every set map T → M to a commutative monoid extends uniquely 100 PETE L. CLARK ⊕ to a homomorphism t∈T Z → M (in other words, we pass from the category of sets to the category of commutative R-algebras by passing through the category of commutative monoids, taking the free commutative monoid associated to a set and then the free R-algebra associated to the monoid). As before, the uniqueness of the extension is easy to verify. Exercise 5.39: a) Formulate analogous universal properties for Laurent polynomial rings, and non-commutative polynomial rings. b) Suppose M is a divisor-finite monoid. Is there an analogous extension property for the big monoid ring R[[M ]]? This result is of basic importance in the study of R-algebras. For instance, let S be an R-algebra. A generating set for S , as an R-algebra, consists of a subset T of S such that the least R-subalgebra of S containing T is S itself. This definition is not very concrete. Fortunately, it is equivalent to the following: Theorem 152. Let R be a commutative ring, S a commutative R-algebra, and T a subset of S . TFAE: (i) T generates S as an R-algebra. (ii) The canonical homomorphism of R-algebras R[T ] → S – i.e., the unique one sending t → t – is a surjection. Exercise 5.40: Prove Theorem 152. In particular, a commutative R-algebra S is finitely generated iff it is a quotient ring of some polynomial ring R[t1 , . . . , tn ]. Another application is that every commutative ring whatsoever is a quotient of a polynomial ring (possibly in infinitely many indeterminates) over Z. Indeed, for a ring R, there is an obvious surjective homomorphism from the polynomial ring Z[R] – here R is being viewed as a set of indeterminates – to R, namely the one carrying r → r. A ring R is said to be absolutely finitely generated if it is finitely generated as a Z-algebra; equivalently, there exists an n ∈ N and an ideal I in Z[t1 , . . . , tn ] such that Z[t1 , . . . , tn ] is isomorphic to R. Exercise 5.41: a) Show that every finitely generated ring has finite or countably infinite cardinality. b) Find all fields which are finitely generated as rings. (N.B.: In field there is another notion of absolute finite generation for a field. This a much weaker notation: e.g. Q(x) is absolutely finitely generated as a field but not as a ring.) 6. Swan’s Theorem We now digress to discuss an important theorem of R.G. Swan on projective modules over rings of continuous functions. Throughout this section K denotes either the field R or the field C, each endowed with their standard Euclidean topology. For a topological space X , the set C (X ) COMMUTATIVE ALGEBRA 101 of all continuous functions f : X → K forms a commutative ring under pointwise addition and multiplication. 6.1. Introduction to (topological) vector bundles. Recall25 the notion of a K-vector bundle over a topological space X . This is given by a topological space E (the “total space”), a surjective continuous map π : E → X and on each fiber Ex := π −1 (x) the structure of a finite-dimensional K -vector space satisfying the following local triviality property: for each x ∈ X , there exists an open neighborhood U containing x and a homeomorphism f : π −1 U → U × K n such that for all y ∈ U f carries the fiber Ey over y to {y } × K n and induces on these fibers an isomorphism of K -vector spaces. (Such an isomorphism is called a local trivialization at x.) As a matter of terminology we often speak of “the vector bundle E on X ” although this omits mention of some of the structure. On any K -vector bundle E over X we have a rank function r : X → N, namely we define r(x) to be the dimension of the fiber Ex . We say that E is a rank n vector bundle if the rank function is constantly equal to n. The existence of local trivializations implies that the rank function is locally constant – or equivalently, continuous when N is given the discrete topology, so if the base space X is connected the rank function is constant. As a basic and important example, for any n ∈ N we have the trivial rank n vector bundle on X , with total space X × K n and such that π is just projection onto the first factor. If π : E → X and π ′ : E ′ → X are two vector bundles over X , a morphism of vector bundles f : E → E ′ is a continuous map of topological spaces from E to E ′ over X in the sense that π = π ′ ◦ f – equivalently f sends the fiber Ex to the ′ fiber Ex – and induces a K -linear map on each fiber. In this way we get a category Vec(X ) of K -vector bundles on X . If we restrict only to rank n vector bundles and morphisms between them we get a subcategory Vecn (X ). A vector bundle E on X is said to be trivial (or, for emphasis, “globally trivial”) if it is isomorphic to the trivial rank n vector bundle for some n. Many of the usual linear algebraic operations on vector spaces extend immediately to vector bundles. Most importantly of all, if E and E ′ are two vector bundles on X , we can define a direct sum E ⊕ E ′ , whose definining property is that its fiber over each point x ∈ X is isomorphic to Ex ⊕ Ex′ . This not being a topology/geometry course, we would like to evade the precise construction, but here is the idea: it is obvious how to define the direct sum of trivial bundles. So in the general case, we define the direct sum by first restricting to a covering family {Ui }i∈I of simultaneous local trivializations of E and E ′ and then glue together these vector bundles over the Ui ’s. In a similar way one can define the tensor product E ⊗ E ′ and the dual bundle E ∨ . For our purposes though the direct sum construction is the most important. It 25from a previous life, if necessary 102 PETE L. CLARK gives Vec(X ) the structure of an additive category: in addition to the existence of direct sums, this means that each of the sets Hom(E, E ′ ) of morphisms from E to E ′ form a commutative group. (In fact Hom(E, E ′ ) naturally has the structure of a K -vector space.) Decategorifying, the set of all isomorphism classes of vector bundles on X naturally forms a commutative monoid under direct sum (the identity is the trivial vector bundle X → X where each one point fiber is identified – uniquely! – with the zero vector space). The Grothendieck group of this monoid is K (X ): this is the beginning of topological K-theory. 6.2. Swan’s Theorem. But we digress from our digression. A (global) section of a vector bundle π : E → X is indeed a continuous section σ of the map π , i.e., a continuous map σ : X → E such that π ◦ σ = 1X . The collection of all sections to E will be denoted Γ(E ). Again this is a commutative group and indeed a K -vector space, since we can add two sections and scale by elements of K . But in fact more is true. The global sections form a module over the ring C (X ) of continuous K -valued functions, in a very natural way: given a section σ : X → E and f : X → K , we simply define f σ : X → E by x → f (x)σ (x). Thus Γ : E → Γ(E ) gives a map from vector bundles over X to C (X )-modules. In fancier language, Γ gives an additive functor from the category of vector bundles on X to the category of C (X )-modules; let us call it the global section functor. (Indeed, if we have a section σ : E → X of E and a morphism of vector bundles f : E → E ′ , f (σ ) = f ◦ σ is a section of E ′ . No big deal!) Theorem 153. (Swan [Sw62]) Let X be a compact space. Then the global section functor Γ gives an equivalence of categories from Vec(X ) to the category of finitely generated projective C (X )-modules. In other words, at least for this very topologically influenced class of rings C (X ), we may entirely identify finitely generated projective bundles with a basic and important class of geometric objects, namely vector bundles. There is a special case of this result which is almost immediately evident. Namely, suppose that E is a trivial vector bundle on X , i.e., up to isomorphism E is simply X × K n with π = π1 . Thus a section σ is nothing else than a continuous function σ : X → K n , which in turn is nothing else than an n-tuple (f1 , . . . , fn ) of elements of C (X ). Thus if we define σi ∈ Γ(E ) simply to be the section which takes each point to the ith standard basis vector ei of K n , we see immediately that (σ1 , . . . , σn ) is a basis for Γ(E ), i.e., Γ(E ) is a free C (X )-module of rank n. Moreover, we have Hom(X × K n , X × K m ) ∼ Map(X, HomK (K n , K m )) = ∼ C (X ) ⊗K Hom(K n , K w ) ∼ HomC (X ) (Γ(X × K n ), Γ(X × K m )). = = Thus we have established that Γ gives an additive equivalence from the category of trivial vector bundles on X to the category of finitely generated free C (X )-modules. We wish to promote this to an equivalence from locally trivial vector bundles (i.e., all vector bundles) to finitely generated projective modules. Oh, if only we had some nice “geometric” characterization of finitely generated projective modules! COMMUTATIVE ALGEBRA 103 But we do: namely Proposition 19 characterizes finitely generated projective modules over any commutative ring R as being precisely the images of projection operators on finitely generated free modules. Thus the essence of what we want to show is that for any vector bundle E over X (a compact space), there exists a trivial vector bundle T and a projection P : T → T – i.e., an element of Hom(T, T ) with P 2 = P such that the image of P is a vector bundle isomorphic to E . Indeed, if we can establish this, then just as in the proof of 19 we get an internal direct sum decomposition T = P (T ) ⊕ (1 − P )(T ) and an isomorphism P (T ) ∼ E , and = applying the additive functor Γ this gives us that Γ(E ) is isomorphic to a direct summand of the finitely generated free C (X )-module Γ(T ). A little thought shows that in fact this proves the entire result, because we have characterized Vec(X ) as the “projection category” of the additive category trivial vector bundles, so it must be equivalent to the “projection category” of the equivalent additive category of finitely generated free C (X )-modules. So from this point on we can forget about projective modules and concentrate on proving this purely topological statement about vector bundles on a compact space.26 6.3. Proof of Swan’s Theorem. Unfortunately the category of vector bundles over X is not an abelian category. In particular, it can happen that a morphism of vector bundles does not have either a kernel or image. Swan gives the following simple example: let X = [0, 1], E = X × K the trivial bundle, and f : E → E be the map given by f (x, y ) = (x, xy ). Then the image of f has rank one at every x ̸= 0 but has rank 0 at x = 0. Since X is connected, a vector bundle over X should have constant rank function. Exactly the same considerations show that the kernel of f is not a vector bundle. However, nothing other than this can go wrong, in the following sense: Proposition 154. For a morphism f : E → E ′ of vector bundles over X , TFAE: (i) The image of f is a subbundle of E ′ . (ii) The kernel of f is a subbundle of E . (iii) The function x → dimK (Im f )x is locally constant. (iv) The function x → dimK (Ker f )x is locally constant. Proof. Step 1: We first wish to prove a special case: namely that if f : E → E ′ is a monomorphism of vector bundles (i.e., it induces an injection on all fibers) then (Im f ) is a subbundle of E ′ and f : E → (Im f ) is an isomorphism. The issues of whether Im f is a vector bundle and f is an isomorphism are both local ones, so it suffices to treat the case where E and E ′ are trivial bundles. Suppose E ′ = X × V , and let x ∈ X . Choose Wx ⊂ V a subspace complementary to (Im f )x . Then G := X × Wx is a sub-bundle of E ; let ι : G → E be the inclusion map. Define θ : E ⊕ G → E ′ by θ((a, b)) = f (a) + ι(b). Then θx is an isomorphism, so there exists an open neighborhood U of x such that θ|U is an isomorphism. Since E is a subbundle of E ⊕ G, θ(E ) = f (E ) is a subbundle of θ(E ⊕ G) = E ′ on U . Step 2: Since the rank function on a vector bundle is locally constant, (i) =⇒ 26We note that [Sw62] takes a more direct approach, for instance proving by hand that the global section functor Γ is fully faithful. In our use of projection operators and projection categories to prove Swan’s theorem we follow Atiyah [At89, §1.4]. Aside from being a bit shorter and slicker, this approach really brings life to Proposition 19 and thus seems thematic in a commutative algebra course. But it is not really more than a repackaging of Swan’s proof. 104 PETE L. CLARK (iii), (ii) =⇒ (iv), and (by simple linear algebra!) (iii) ⇐⇒ (iv). (iv) =⇒ (i): Again the issue of whether Im f is a vector bundle is a local one, so we may assume that E = X × V is a trivial bundle. For x ∈ X , let Wx ⊂ V be a complementary subspace to (Ker f )x . Let G = X × Wx , so that f induces a homomorphism ψ : G → E ′ whose fiber at x is a monomorphism. Thus ψ is a monomorphism on some neighborhood U of x, so ψ (G)|U is a subbundle of E ′ |U . However Ψ(G) ⊂ f (E ), and since f (E ) has constant rank, and dim ψ (G)y = dim ψ (G)x = dim f (E )x = dim f (E )y for all y ∈ U , ψ (G)|U = f (E )|U . so f (E ) is a subbundle of E ′ . (iv) =⇒ (ii): here we exploit dual bundles. The hypothesis implies that the kernel of f ∨ : (E ′ )∨ → E ∨ has constant rank function. Since E ∨ → Coker f ∨ is an epimorphism, (Coker f ∨ )∨ → E ∨∨ is a monomorphism: by Step 1, its image is a subbundle. But the natural map E → E ∨∨ is an isomorphism, the restriction of which to Ker f gives an isomorphism to the vector bundle (Coker f ∨ )∨ . So Ker f is a vector bundle. The proof yields the following additional information. Corollary 155. For any morphism of vector bundles, the rank function of the image is upper semi-continuous: that is, for any x ∈ X , there exists a neighborhood U of x such that for all y ∈ U , dimK (Im f )y ≥ dimK (Im f )x . Exercise 6.1: Prove Corollary 155. Proposition 156. Let E be a vector bundle over X , and let P ∈ End(E ) = Hom(E, E ) be a projection, i.e., P 2 = P . Then: a) We have Ker(P ) = Im(1 − P ). b) Im P and Im(1 − P ) are both subbundles of E . c) There is an internal direct sum decomposition E = Im P ⊕ Im(1 − P ). Proof. a) For all x ∈ X linear algebra gives us an equality of fibers Ker(P )x = Im(1 − P )x . This suffices! b) From part a) we deduce an equality of rank functions rIm P + rIm(1−P ) = rE . By Corollary 155, for all x ∈ X , there is a neighborhood U of x on which rIm P is at least as large as rIm P (x), rIm(1−P ) is at least as large as rIm(1−P ) (x) and rE is constantly equal to rE (x). On this neighborhood the ranks of Im P and Im(1 − P ) must be constant, and therefore by Proposition 154 Im P and Im(1 − P ) are both subbundles. c) Again it is enough to check this fiber by fiber, which is simple linear algebra. An inner product on a finite-dimensional R-vector space V is, as usual, a symmetric R-bilinear form ⟨, ⟩ : V × V → R which is positive definite in the sense that for all x ∈ V \ {0}, ⟨x, x⟩ = 0. An inner product on a finite-dimensional C-vector space V is a positive definite sesquilinear form: i.e., it is C-linear in the first variable, conjugate-linear in the second variable and again we have ⟨x, x⟩ > 0 for all x ∈ V \ {0}. Now let E be a K -vector bundle on X . An inner product on E is a collection of inner products ⟨, ⟩x : Ex × Ex → K on each of the fibers which vary continuously in COMMUTATIVE ALGEBRA 105 x. Formally this means the following: let E ×X E be the subset of (e1 , e2 ) ∈ E × E such that π (e1 ) = π (e2 ); then such a fiberwise family of inner products defines a function from E ×X E to K , and this function is required to be continuous. Let us say that a metrized vector bundle E on X is a vector bundle together with an inner product. (Again, this is an abuse of terminology: we do not speak of the inner product by name.) Proposition 157. Let E be a metrized line bundle on X . a) If E ′ is a subbundle of E , fiberwise orthogonal projection onto E ′ defines a projection operator P ∈ End(E ) with image E ′ . b) All short exact sequences 0 → E ′ → E → E ′′ → 0 of vector bundles are split. c) If M is another vector bundle on X and there exists an epimorphism of bundles q : E → M , then M is isomorphic to the image of a projection operator on E . Proof. a) This is mostly a matter of understanding and unwinding the definitions, and we leave it to the reader. b) Let P be orthogonal projection onto E ′ . The restriction of the map E → E ′′ to Ker P is an isomorphism of vector bundles. The inverse of this isomorphism gives a splitting of the sequence. c) By Proposition 154, since Im q = M is a vector bundle, so is Ker q , whence a short exact sequence 0 → Ker q → E → M → 0. By part b), there exists a splitting σ : M → E of this sequence. Then, as usual, ∼ P = σ ◦ q is a projection operator on E and q |Im P : Im P → M . Proposition 158. If X is a paracompact topological space, then every vector bundle over X admits an inner product. Proof. This is a rather standard topological argument which we just sketch here. Let M be a vector bundle on X , and let {Ui }i∈I be an open covering of X such that the restriction of M to each Ui is a trivial bundle. On a trivial bundle there is an obvious inner product, say ⟨, ⟩x . Now, since X is paracompact, there exists a partition of unity {φi }i∈I subordinate to the open covering {Ux }: that is, • each φi : X → [0, 1] is continuous, • for all x ∈ X we have supp(φi ) ⊂ Ui • for all x ∈ X there exists an open neighborhood V of x on which all but finitely many φi ’s vanish identically, and ∑ • for all x ∈ X , i∈I φi (x) = 1.27 Then, for x ∈ X and e1 , e2 ∈ Mx , define ∑ ⟨e1 , e2 ⟩x := φi (x)⟨e1 , e2 ⟩i ; i the sum extends over all i ∈ I such that x ∈ Ui . This is an inner product on M . To complete the proof of Swan’s Theorem, it suffices to show that if X is compact, every vector bundle M on X is the epimorphic image of a trivial bundle. In particular, Proposition X.X then shows that M is a direct summand of a trivial vector bundle T and thus Γ(M ) is a direct summand of the finitely generated free C (X )-module Γ(T ), hence is finitely generated projective. 27See e.g. Exercise 5 in §4.5 of Munkres’ Topology: a first course for a proof of this fact. 106 PETE L. CLARK Proposition 159. Let X be a compact space and M a vector bundle on X . Then there exists an epimorphism of bundles from a trivial vector bundle X × V to M . Proof. Step 1: We claim that for each x ∈ X , there exists a neighborhood Ux of x and finite set of global sections Sx = {sx,1 , . . . , sx,kx } of M such that for all y ∈ U , sx,1 (y ), . . . , sx,kx (y ) is a K -basis for My . proof of claim: Let U be an open neighborhood of x on which M is a trivial bundle. Certainly then there exist finitely many sections s1 , . . . , sn of M over U which when evaluated at any y ∈ U give a basis of My . We need to show that there exists an open set W with x ∈ W ⊂ U and global sections s′ , . . . , s′ such that n 1 for all i, s′ |W = si |W . For this it suffices to work one section at a time: let s be i a section of M over U . Since X is paracompact, it is normal, so there exist open neighborhoods W and V of x with W ⊂ V , V ⊂ U . By Urysohn’s Lemma, there is a continuous function ω : X → [0, 1] such that ω |W ≡ 1 and ω |X \V ≡ 0. If we then define s′ : X → M by s′ (y ) = ω (y )s(y ) for y ∈ U and s′ (y ) = 0 for y ∈ X \ U , then this s does the job. Step 2: By compactness of X , there exists a finite subset I of X such that {Ux }x∈I ∪ covers X . So S = i∈I Sx is a finite set of global sections of M which when evaluated at any x ∈ X , span the fiber Mx . So the K -subspace V of Γ(M ) spanned by S is finite-dimensional. We define a map q : X × V → M by q (x, s) = s(x). This is a surjective bundle map from a trivial vector bundle to M ! Remark: In the above proof the paracompactness of X seems to have been fully exploited, but the need for compactness is less clear. In fact, at the end of [Sw62], Swan remarks that if you replace the last step of the proof by an argument from Milnor’s 1958 lecture notes Differential Topology, one gets a categorical equivalence between vector bundles with bounded rank function on a paracompact space X and finitely generated projective C (X )-modules. Remark: A more straightforward variant of Swan’s theorem concerns the case where X is a compact differentiable manifold (say of class C ∞ ). In this case the equivalence is between differentiable K -vector bundles on X and modules over the ring of K -valued C ∞ -functions. Looking over the proof, one sees that the only part that needs additional attention is the existence of differentiable partitions of unity. Such things indeed exist and are constructed in many of the standard texts on geometry and analysis on manifolds. We recommend [Wel80], which has a particularly clear and complete discussion. 6.4. Applications of Swan’s Theorem. 6.4.1. Vector bundles and homotopy. Vector bundles on a space are of interest not only to differential topologists and geometers but also to algebraic geometers. This is because pullback of vector bundles behaves well under homotopy. First, suppose that f : X → Y is a continuous map of topological spaces and π : E → Y is a vector bundle on Y . We may pullback π to a vector bundle πX : E ×Y X → X just by taking E ×Y X to be the fiber product of the maps f and π , namely the subspace of X × E consisting of all pairs (x, v ) such that COMMUTATIVE ALGEBRA 107 f (x) = π (v ) ∈ Y . The map πX : E ×Y X → X is just restriction of the projection map: (x, v ) → x. Exercise 6.2: Show that πX : E ×Y X → X is indeed a vector bundle on X . We abbreviate it by either f ∗ π or (more abusively) f ∗ E . Exercise 6.3: Show that the pullback of any trivial bundle is a trivial bundle. Theorem 160. (Covering Homotopy Theorem) Let X and Y be topological spaces with X paracompact. Let π : E → Y be a vector bundle on Y , and let f, g : X → Y be homotopic maps. Then the pullbacks f ∗ π and g ∗ π are isomorphic vector bundles on X . Proof. See for instance [Hus66, Thm. 4.7]. For our applications, it is enough to know that compact spaces are paracompact. But for culture we also remark that any regular σ -compact space is paracompact, e.g. any CW-complex with only finitely many cells of any given dimension. Corollary 161. If X is a contractible paracompact space, then every vector bundle on X is trivial. Proof. Choose any point x0 ∈ X , let f : X → X be the map which sends every point of X to x0 ,and let g : X → X be the identity map. If π : E → X is any vector bundle on X , then by Theorem 160 we have f ∗ π = g ∗ π . Since g is the identity map, g ∗ π = π . On the other hand, tracking through the definitions shows f ∗ π = X × π −1 (x0 ), a trivial bundle. So π is trivial. 6.4.2. Projective Modules over C ([0, 1]). Let [0, 1] be the closed unit interval with its topology: a compact, contractible space. By Corollary 161, every vector bundle over [0, 1] is trivial. By Swan’s Theorem, this implies that every finitely generated projective module over the ring R = C ([0, 1]) of continuous functions f : [0, 1] → R is free. But now – as in §3.9 – consider the ideal I of all functions f ∈ R which vanish near zero, i.e., for which there exists ϵ(f ) > 0 such that f |[0,ϵ(f )] ≡ 0. By Exercise X.X, I is a projective R-module. Moreover, I is not a free R-module: indeed, any f ∈ I is annihilated by any continuous function with support lying in [0, ϵ(f )], and nonzero such functions clearly exist. On the other hand, any nonzero free module has elements with zero annihilator: take any basis element. Thus C ([0, 1]) is a connected ring over which every finitely generated projective module is free, but the infinitely generated projective module I is not free. (Recall that Theorem 25 says that no such modules exist over connected Noetherian rings.) Moreover I is therefore clearly not a direct sum of finitely generated modules, since by what we have established any such module over C ([0, 1]) would be free! Exercise 6.4: Use Corollary 56 to give a purely algebraic proof that I is not a direct sum of finitely generated submodules. 108 PETE L. CLARK Exercise 6.5*: Find necessary and sufficient conditions on a compact, contractible space X for there to exist a nonfree projective module. 6.4.3. Some stably free modules. For n ∈ N, define Rn := R[x0 , . . . , xn ]/(x2 + . . . + x2 − 1). 0 n N Recalling that every Rn -module map from Rn → RM uniquely corresponds to a n matrix M ∈ MM,N (Rn ), consider the map H : Rn+1 → Rn associated to the matrix (x0 · · · xn ). The image of H is an ideal of Rn . Applying H to each of the standard n basis vectors e0 , . . . , en of Rn+1 we see that x0 , . . . , xn all lie in the image of H , 2 2 hence so does x0 + . . . + xn = 1: that is, H is surjective. We define Pn = ker(H ), so we have a short exact sequence H n 0 −→ Pn −→ Rn+1 −→ Rn −→ 0. Since R is free, this sequence splits: indeed, in this case we have a canonical splitting n given by the section σ : Rn → Rn+1 with associated matrix (x0 · · · xn )t . Thus ∼ Pn ⊕ Rn = Rn+1 , n so Pn is projective and moreover stably free (recall a module M is stably free if there exist finitely generated free modules F1 and F2 with M ⊕ F1 ∼ F2 ). It is = natural to ask whether Pn is in fact a free Rn -module. Here is the answer. Theorem 162. (Swan) The stably free Rn -module Pn is free iff n = 0, 1, 3 or 7. Let us sketch the proof of this theorem. Step 0: To show that a module is free, it is of course enough to find a basis. Note that P0 = 0, so this is a trivial case. (It will become clear later why this case was included.) When n = 1, it is an easy direct calculation to show that −x1 e0 + x0 e1 is a basis for P1 . Similarly concrete formulas can be found for bases of Pn when n = 3 and n = 7. We leave this as an exercise for the reader. Step 1: Now suppose that n is not 0, 1, 3 or 7. We wish to show that Pn is not free. For this it is sufficient to find some extension ring S of R such that Pn ⊗R S is not free. So now is a good time to observe that Rn is nothing else than the ring of polynomial functions on the unit n-sphere S n inside Euclidean space Rn+1 . That is, every polynomial f ∈ R[x0 , . . . , xn ] defines by restriction a real-valued function on S n , but we wish to regard two such polynomials as giving the same function if they agree at every point of S n , but the ideal of functions vanishing identically on S n is precisely I = (x2 + . . . + x2 − 1).28 Since every polynomial function is continuous n 0 in the usual Euclidean topology on S n , we get an extension of rings Rn ⊂ C (S n ). So it is enough to show that the finitely generated projective C (S n )-module Tn := Pn ⊗Rn C (S n ) is not free. 28This part of the argument will become more clear when we study affine rings and the Nullstellensatz later on. Indeed the polynomial x2 + . . . + x2 − 1 is irreducible when n > 0 so the ideal n 0 I is prime, thus radical, so I (V (I )) = I . COMMUTATIVE ALGEBRA 109 Step 2: Now we are in a Swan’s Theorem situation: the module Tn corresponds to a unique vector bundle on S n . In fact we claim that it corresponds to the tangent bundle T S n on S n . Since we have not given a rigorous definition or construction of the tangent bundle to a manifold, we are not in a position to prove this here, but here are some hints in this direction. We have S n ⊂ Rn+1 . The tangent bundle to Rn+1 is of course trivial, hence so is its pullback – here, restriction – to S n , say F n+1 . Further, there is a surjective bundle map from F n+1 to the trivial bundle of rank one, say F 1 : at every point of the n-sphere we take the outward unit normal vector. The kernel of this bundle map is the tangent bundle to S n . Thus we have a short exact sequence of vector bundles: 0 −→ T S n −→ F n+1 −→ F 1 −→ 0. We claim that under the equivalence of Swan’s Theorem, this split exact sequence corresponds to the split exact sequence H 0 −→ Tn −→ C (S n )n+1 −→ C (S n ) −→ 0, the base change to C (Sn ) of the short exact sequence of Rn -modules defining Pn . Step 3: We invoke the following classical theorem. Theorem 163. (Bott-Milnor [BM58]) The tangent bundle to the n-sphere is trivial iff n = 0, 1, 3 or 7. This completes the proof of Theorem 162! Theorem 163 is of course a deep and celebrated result: the original proof used recently developed properties of Stiefel-Whitney classes, Pontrjagin classes, cohomology operations, and so forth. In 1962 J.F. Adams solved a more general problem: for each n he found the largest rank k of a trivial subbundle of the tangent bundle to S n [Ad62]. Somewhat more recently it was found that K-theory gives more graceful proofs: we recommend that the reader consult for instance [Ka08, §V.2]. If one merely wants some values of n for which Pn is not free, one can use much lower-powered differential topological machinery. Indeed, the Poincar´-Hopf theoe rem [Mi65, p. 35] shows that a closed n-manifold can only admit an everywhere nonvanishing vector field (this is much weaker than triviality of its tangent bundle, which involves finding n everywhere linearly independent vector fields) if its Euler characteristic is 0. But the Euler characteristic of S n is 1 + (−1)n , so is nonzero for all even n. Thus Bott-Milnor is not needed to show that Pn is nonfree for all even n. One can of course also ask whether all this topological stuff is really necessary to prove the nonfreeness of the stably free modules Pn . The answer is: apparently, yes! So far as I know, there is no purely algebraic proof of Theorem 162. 7. Localization 7.1. Definition and first properties. As we have seen, one way to “simplify” the study of ideals in a ring R is to pass to a quotient ring R/I : as we have seen, this has the (often useful) effect of “cutting off the bottom” of the ideal lattice by keeping only ideals J ⊃ I . There is another 110 PETE L. CLARK procedure, localization, which effects the opposite kind of simplification: given a prime ideal P of R, there is a ring RP together with a canonical map ι : R → RP such that ι∗ : I (RP ) → I (R) is an injection whose image is precisely the ideals J ⊂ P . As usual, ι∗ carries prime ideals to prime ideals. In particular, assuming only that P is prime, we get a corresponding ideal – rather inelegantly but standardly denoted P RP – which is the unique maximal ideal of RP . If we can take P = (0) – i.e., if R is a domain – this means that P RP is the only ideal of RP , which is therefore a field. In fact it is nothing else than the quotient field of the integral domain R, and – with one exception – all the secrets of localization are already present in this very familiar special case. In fact the localization construction is a bit more general than this: given an arbitrary ring R (of course commutative with unity!) and an arbitrary multiplicative subset S of R – this just means that 1 ∈ S and SS ⊂ S – we will define a new ring RS together with a canonical homomorphism ι : R → RS (for which ι∗ will still be an injection with explicitly given image). In fact, just as in the case of quotients, ι satisfies a certain universal mapping property, but let us sacrifice some elegance for intelligibility by working our way up to this crisp definition. Indeed, first consider the special case in which R is a domain, with fraction field F . Then RS will be an extension ring of R, still with fraction field F , which is obtained by adjoining to R all elements 1 for s ∈ S . s Example: Suppose R = Z, S = {2n }n∈N . Then RS = Z[ 1 ]. Indeed we see that for 2 any nonzero element f , we can take S to be the multiplicative set consisting of the 1 powers of f , and then the localization is just R[ f ]. What if in the example above, instead of taking the multiplicative subset generated by 2, we took the multiplicative subset generated by 22 , or 2127 ? Clearly it k −1 wouldn’t matter: if we have 21 for any k in our subring of Q, we also have 2 2k = 1 . k 2 To generalize this idea, define the saturation S of a multiplicatively closed subset S of a domain R to be the set {a ∈ R | ∃b ∈ R | ab ∈ S }, i.e., the set of all divisors of elements of S . The same observation as above shows that RS = RS , so if we like we can restrict to consideration of saturated multiplicatively closed subsets. Example, continued: The saturated, multiplicatively closed subsets of Z correspond to (arbitary) subsets P of the prime numbers (exercise!). In particular Z 1 itself corresponds to P = ∅, Z[ p ] corresponds to P = {p}, Q corresponds to the set of all primes. Most interestingly, fix any prime p and let P be the set of all primes except p: then the corresponding ring, which is confusingly denoted Z(p) is the set of all rational numbers of the form x where p does not divide y . Notice that y such rings are the maximal subrings of Q which are not fields. Moreover, the units of Z(p) are precisely the elements of the form x with (p, x) = 1. The nonunits a y are all of the form pa′ for a′ ∈ Z(p) , so therefore the unique maximal ideal is the principal ideal (p) = pZ(p) . Exercise 7.1: Show that the only ideals in Z(p) are those of the form (p)k for some k ∈ N. Notice that this set happens to be identifiable with the set of all ideals COMMUTATIVE ALGEBRA 111 I of Z which are disjoint from the multiplicative set S (P ). Now let R be any ring and S a multiplicatively closed subset of R. We would still like to define a ring S −1 R which is, roughly speaking, obtained by adjoining to R all inverses of elements of S . We can still define S −1 R in terms of formal quotients, i.e., as equivalence classes of elements (a, b) with a ∈ R, b ∈ S . However, if we define (a, b) ∼ (c, d) to be ad = bc, then unfortunately we find that this need not be an equivalence relation! Therefore we need to enlarge the relation a bit: we put (a, b) ∼ (c, d) iff there exists s ∈ S such that sad = sbc. We then define ab at + bs + := , s t st ab ab · := . st st We must check that these operations are well-defined on equivalence classes; this is left as a (perhaps somewhat tedious, but not difficult) exercise for the reader. Exercise 7.2: Indeed, check that S −1 R is a ring and that x → x defines a ho1 momorphism of rings R → S −1 R. Thus S −1 R is an R-algebra, and in particular an R-module. Exercise 7.3: Let R be a domain and S = R• = R \ {0}. Show that S −1 R is indeed the fraction field of R. When f ∈ R, we denote the localization of R at the multiplicative subset generated by f as Rf . Example: Suppose f ∈ R is a nilpotent element: f n = 0 for some n ∈ Z+ . Then f n−1 0 1 = f n−1 whereas 0 = f . Since (f n−1 · f − f n−1 · 0) = 0, we have that 1 = 0, i.e., Rf is the zero ring. Conversely, if f is not nilpotent, then if it is a unit, Rf = R. 1 Otherwise, all the powers of f are distinct, and then f ̸= 1 , since for any n ∈ N, 1 n f (1 − f ) ̸= 0, so that Rf is not the zero ring. In general, S −1 R is the zero ring iff S contains 0. This is to be regarded as a trivial case, and may safely be tacitly excluded in the sequel. Exercise 7.4: a) Show that the kernel of the natural map R → S −1 (R) is the set of all r ∈ R such that for some s ∈ S , sr = 0. b) The map R → S −1 (R) is injective iff S has no zerodivisors. c) Show that the subset Q of all nonzerodivisors of a ring R is multiplicatively closed. The localization Q−1 R is called the total fraction ring of R. Show that Q−1 (R) is a field iff R is an integral domain. Exercise 7.5: Show that the homomorphism R → S −1 R is universal for homomorphisms R → T with f (S ) ⊂ T × . 7.2. Pushing and pulling via a localization map. Let R be a ring and S a multiplicatively closed subset. Let ι : R → S −1 R be 112 PETE L. CLARK the natural map. As for any homomorphism of rings, ι induces maps between the sets of ideals of R and the set of ideals of S −1 R, in both directions: ι∗ : IR → IS −1 R , I → IS −1 R, ι∗ : IS −1 R → IR , J → ι−1 (J ). Lemma 164. Let ι : R → S −1 R be a localization map. Then for any ideal I of R, x ι∗ (I ) = { ∈ S −1 R | x ∈ I, s ∈ S }. s Proof. Let us temporarily write x I = { ∈ S −1 R | x ∈ I, s ∈ S }. s We want to show that I = ι∗ (I ) = ⟨ι(I )⟩S −1 R . It is clear that ι(I ) ⊂ I ⊂ ι∗ (I ), so 1 2 it is enough to show that I is itself an ideal of S −1 R. No problem: if x1 , x2 ∈ I , s s x1 x2 x1 s2 + x2 s1 + = ∈ I, s1 s2 s1 s2 and if y s ∈ S −1 R, then y x1 x1 y = ∈ I. s s1 ss1 Like quotient maps, any localization map has the pull-push property. Proposition 165. Let ι : R → S −1 R be a localization. For any ideal J of S −1 R, J = ι∗ ι∗ J. Proof. We have seen before that for any homomorphism ι : R → R′ of rings and any ideal J of R′ we have J := ι∗ ι∗ J ⊂ J. Thus it is enough to show the reverse containment. For this, consider an arbitrary element x ∈ J . Then x = s x ∈ J hence also x ∈ ι∗ (J ), so ι(x) ∈ J . But since J is s s an ideal and s is a unit in S −1 R, we then also have 1 x = x ∈ J . s s Lemma 166. Let ι : R → S −1 R be a localization map and I an ideal of R. TFAE: (i) I ∩ S ̸= ∅. (ii) ι∗ (I ) = S −1 R. Proof. (i) =⇒ (ii): If s ∈ S ∩ I , then s ∈ IS −1 R, so 1 = s ∈ ι∗ (I ). s (ii) =⇒ (i): Suppose 1 ∈ ι∗ (I ). By Lemma 164, 1 = x for some x ∈ I and s ∈ S . 1 s Clearing denominators, there is s′ ∈ S such that ss′ = s′ x and thus ss′ ∈ I ∩ S . Proposition 167. Let ι : R → S −1 R be a localization homomorphism. a) For a prime ideal p of R, TFAE: (i) The pushforward ι∗ p is prime in S −1 R. (ii) The pushforward ι∗ p is proper in S −1 R. (iii) The sets p ∩ S = ∅. b) If p is prime and disjoint from S , then ι∗ (ι∗ p) = p. COMMUTATIVE ALGEBRA 113 Proof. a) By Lemma 166, (ii) ⇐⇒ (iii) for all ideals of R. (i) =⇒ (iii): By contraposition. Once again, if s ∈ p ∩ S , then ι∗ (p) = S −1 R: that is, it is not even proper, let alone prime. 1 2 (iii) =⇒ (i): Suppose p is a prime ideal of R, and suppose we have a1 , a2 ∈ S −1 R s s a1 a2 x ′ with s1 s2 = s ∈ ι∗ (p). Clearing denominators, there is s ∈ S such that ss′ a1 a2 = s′ s1 s2 x ∈ p. Since S ∩ p = ∅, (ss′ ) ∈ p, and since p is prime, we conclude that a1 a2 ∈ p and / i then that ai ∈ p for some i, hence ai ∈ ι∗ p for some i and ι∗ (p) is prime. This s completes the proof of part a). b) Recall: for any homomorphism ι : R → R′ and any ideal I of R we have ι∗ (ι∗ (I )) ⊃ I, so taking I = p to be prime it suffices to show the inverse inclusion. Suppose x ∈ ι∗ ι∗ p, i.e., there exist a ∈ p, s ∈ S such that ι(x) = x = a . By definition, this 1 s means that there exists some s′ ∈ S such that s′ sx = s′ a ∈ p. Therefore either s′ s ∈ p or x ∈ p, but since s′ s ∈ S and S is disjoint from p, we must have x ∈ p. Corollary 168. The maps ι∗ and ι∗ give mutually inverse bijections from the set of prime ideals of S −1 R to the set of prime ideals of R which are disjoint from S . Therefore we may – and shall – view Spec S −1 R as a subset of Spec R. Exercise 7.6: a) Show that the results of Proposition 167 extend to all primary ideals of R.29 b) Let I be any ideal of R. Show that ι∗ ι∗ I = {x ∈ R | ∃s ∈ S such that sx ∈ I }. Comment: in class I remarked that there is no nice push-pull formula for an arbitrary ideal in a localization map. Whether this exercise contradicts that assertion depends upon how nice you find this formula to be! Try it out on the following: c) Exhibit a map ι : R → S −1 R and a (nonprimary) ideal I of R such that ι∗ ι∗ I I . 7.3. The fibers of a morphism. Let f : R → S be a homomorphism of rings, and let p ∈ Spec R. Consider the “fiber of f ∗ : Spec S → Spec R over p”, i.e., fp = (f ∗ )−1 (p) = {P ∈ Spec S | f ∗ (P ) = p}. We claim that fp is canonically isomorphic to the spectrum of a certain ring. Namely, let k (p) be the fraction field of the domain R/p. Then we wish to identify fp with Spec(S ⊗R k (p)). Let ι1 : S → S ⊗R k (p) and ι2 : k (p) ⊗R S ⊗R k (p) be the canonical maps. The tensor product of R-algebras fits into a commutative square (INSERT) and is indeed the categorical pushout: in other words, given any ring A and homomorphisms φ1 : A → S φ2 : A → k (p) such that the composite homomorphisms ι1 ◦ φ1 = ι2 ◦ φ2 are equal, there exists a unique homomorphism Φ : A → R such that f ◦ Φ = φ1 29Recall p is primary if for a, b ∈ R such that ab ∈ p, either a ∈ p or bn ∈ p for some n ∈ Z+ . We have not yet done much with this concept, and will not really address it squarely until the section on primary decomposition. 114 PETE L. CLARK and q ◦ Φ = φ2 . On the spectral side, all the arrows reverse, and the corresponding diagram is (INSERT), which expresses Spec(S ⊗R k (p)) as the fiber product of Spec S and Spec k (p) over Spec R. Observe that the map ι1 : S → S ⊗R k (p) is the composite of the surjective map q1 : S → S ⊗R R/p with the map ℓ2 : S ⊗R R/p → (S ⊗R R/p) ⊗R/p k (p), the latter map being localization with respect to the multiplicatively closed subset q1 (R \ p). ∗ ∗ Both q1 and ℓ∗ are injections, and therefore ι1 ∗ = q1 ◦ ℓ∗ is injective. Similarly 2 2 Spec k (p) → Spec R (this is just the special case of the above with R = S ). It follows that the above diagram identifies Spec S ⊗R k (p) with the prime ideals P of Spec S such that f ∗ P = p. 7.4. Commutativity of localization and passage to a quotient. Lemma 169. Let R be a ring, S ⊂ R a multiplicatively closed subset, and I and ideal of A. Write q : R → R/I for the quotient map and put S := q (S ). Then there is a canonical isomorphism S −1 R/IS −1 R ∼ S = −1 (R/I ). Proof. Explicitly, we send a (mod I )S −1 R to a , where a = a + I , s = s + I . It is s s straightforward to check that this an isomorphism. Remark: Matsumura makes the following nice comment: both sides satisfy the universal property for homomorphisms f : R → R′ such that f (S ) ⊂ (R′ )× and f (I ) = 0. Therefore they must be canonically isomorphic. 7.5. Localization at a prime ideal. An extremely important example of a multiplicative subset of R is the complement R \ p of a prime ideal p. As a matter of notation, we write Rp for (R \ p)−1 R.30 Proposition 170. If p is a prime ideal of R, then the localization Rp is a local ring with unique maximal ideal pRp . Proof. We know that the primes of the localized ring are precisely the pushforwards of the prime ideals of R which are disjoint from the muliplicatively closed set. Here S = R \ p, so being disjoint from S is equivalent to being contained in p. Thus the unique maximal such element is indeed pRp . Remark: We will simply write p for the maximal ideal pRp of Rp . Proposition 170, simple though it is, is of inestimable importance. It shows that the effect of localization at a prime ideal on the lattice of ideals is dual to that of passage to the quotient: if we mod out by a prime p, we get a ring R/p whose ideals are precisely the ideals of R containing p. However, if we localize at R \ p, we get a ring whose ideals are precisely the ideals of R contained in p. In particular, this 30This is inevitably a bit confusing at first, but our choice of notation for a loacalization is designed to make this less confusing. The other common notation for the localization, RS , creates a notational nightmare. As a mnemonic, remember that we gain nothing by localizing at a subset S containing 0, since the corresponding localization is the trivial ring. COMMUTATIVE ALGEBRA 115 construction motivates us to develop an especially detailed theory of local rings, by assuring us that such a theory could be put to good use in the general case. 7.6. Localization of modules. If S is any multiplicative subset and M is any Rmodule, we can also construct a localized R-module S −1 M . One the one hand, we can construct this exactly as we did S −1 R, by considering the appropriate equivalence relation on pairs (m, s) ∈ M × S . On the other hand, we can just take the base extension S −1 R ⊗ M . We are left with the task of showing that these two constructions are “the same”. Exercise 7.7: Formulate a universal mapping property for the localization morphism M → S −1 M . Check that both of the above constructions satisfy this universal mapping property, and deduce that they are canonically isomorphic. Exercise 7.8: Show that the kernel of M → S −1 M is the set of m ∈ M such that ann(m) ∩ S ̸= ∅. Exercise 7.9: Let N be any S −1 R-module. Show that there exists an R-module M such that N ∼ S −1 R ⊗R M . = Generally speaking, thinking of S −1 M as S −1 R ⊗R M is more convenient for proving results, because it allows us to employ the theory of tensor products of modules that we developed in §X.X above. For example: Proposition 171. For any ring R and multiplicatively closed subset S of R, S −1 R is a flat R-module. Equivalently, if 0 → M ′ → M → M ′′ → 0 is a short exact sequence of R-modules, then 0 → S −1 M ′ → S −1 M → S −1 M ′′ → 0 is a short exact sequence of R-modules (or equivalently, of S −1 R-modules). Proof. Tensor products are always right exact, so we need only show S −1 M ′ → ′ S −1 M . Suppose not: then there exists m′ ∈ M ′ and s ∈ S such that m = 0 ∈ M . s ′ Thus there is g ∈ S such that gm′ = 0, but if so, then m = 0 in M ′ .31 s Corollary 172. Let N and P be submodules of an R-module M . Then: a) S −1 (N + P ) = S −1 N + S −1 P . b) S −1 (N ∩ P ) = S −1 N ∩ S −1 P . c) S −1 (M/N ) ∼S −1 R S −1 M/S −1 N . = Exercise 7.10: Prove Corollary 172. Proposition 173. Let M and N be R-modules and S a multiplicatively closed subset of R. Then the mapping m⊗n mn ⊗→ s t st 31Note also that the exactness of a sequence of R-modules does not depend on the R-module structure but only on the underlying abelian group structure. Thus if we have a sequence of abelian groups which can be viewed as a sequence of R-modules and also as a sequence of R′ -modules, then exactness as R-modules is equivalent to exactness as R′ -modules. 116 PETE L. CLARK induces an isomorphism of S −1 (R)-modules ∼ S −1 M ⊗S 1 R S −1 N → S −1 (M ⊗R N ). In particular, for any prime ideal p of R, we have ∼ Mp ⊗Rp Np → (M ⊗R N )p . Exercise 7.11: Prove Proposition 173. Exercise 7.12: Let R be a ring, S a multiplicative subset of R, and M an Rmodule. a) If M is finitely generated, then S −1 M is a finitely generated S −1 R-module. b) If M is finitely presented, then S −1 M is a finitely presented S −1 R-module.32 7.7. Local properties. We say that a property P of a ring R is localizable if whenever R satisfies property P , so does Rp for every prime ideal p of R. We say that a property P is local-to-global if whenever Rp has property P for all prime ideals p of R, then R has that property. Finally, we say a property is local if it is both localizable and local-to-global. There are similar definitions for properties of R-modules. One of the most important themes in commutative algebra is the recognition of the importance of local properties for rings and modules. Remark: Very often it is true that if P is a local property, then R has property P iff Rm has property P for all maximal ideals m of R. We will not introduce terminology for this, but watch for it in the upcoming results. First of all, for an R-module, being trivial is a local property. Proposition 174. For an R-module M , TFAE: (i) M = 0. (ii) Mp = 0 for all primes p of R. (iii) Mm = 0 for all maximal ideals m of R. Proof. Clearly (i) =⇒ (ii) =⇒ (iii), so asume that Mm = 0 for all maximal ideals m of R. Suppose there exists 0 ̸= x ∈ M , and let I be the annihilator of x, so that I is a proper ideal of R and thus contained in some maximal ideal m. Then x is not killed by any element of the multiplicative subset R \ m and therefore maps to a nonzero element of Mm : contradiction. Proposition 175. Let f : M → N be an R-module homomorphism. a) TFAE: (i) f is injective. (ii) For all prime ideals p of R, fp : Mp → Np is injective. (iii) For all maximal ideals m of R, fm : Mm → Nm is surjective. b) Part a) holds with “injective” replaced everywhere by “surjective”, and thus also if “injective” is replaced everywhere by “is an isomorphism.” 32Actually both parts hold for any base change R → R′ ! We record it in this form since it will be used later. COMMUTATIVE ALGEBRA 117 Proof. a) (i) =⇒ (ii) by the exactness of localization, and obviously (ii) =⇒ (iii). Assume (iii), and let M ′ = Ker(f ). Then 0 → M ′ → M → N is exact, hence for ′ ′ all m we have 0 → Mm → Mm → Nm is exact. So, by our assumption, Mm = 0 for ′ all maximal ideals m, and thus by Proposition 174 we have M = 0. The proof of part b) is virtually identical and left to the reader. Warning: Note that Proposition 175 does not say the following: if M and N are R-modules such that Mp ∼ Np as Rp modules for all p ∈ Spec R, then M ∼ N . = = This is being asserted only when there is a map f : M → N inducing all the isomorphisms between localized modules. Exercise 7.13: Exhibit finitely generated R-modules M and N which are “locally isomorphic” – i.e., Mp ∼ Np for all p ∈ Spec R – but are not isomorphic.33 = Corollary 176. Let R be an integral domain with fraction field K . Then as m ∩ ranges over all maximal ideals of R, m Rm = R. ∩ Proof. Consider the injection f : R → S := m Rm . Let p and q be distinct maximal ideals. Then Rq ⊗R Rp = K , so for any maximal ideal m, Sm = Rm and the localized map fm : Rm → Rm is an isomorphism. Therefore f itself is an isomorphism, i.e., surjective. 7.7.1. Local nature of flatness. Proposition 177. For an R-module M , TFAE: (i) M is flat. (ii) For all prime ideals p of R, Mp is flat. (iii) For all maximal ideals m of R, Mm is flat. Proof. (i) =⇒ (ii) is a special case of Proposition 171; (ii) =⇒ (iii) is immediate. So assume (iii), and let N → P be any injective R-module homomorphism. Then, by exactness of localization, for all maximal ideals m we have Nm → Pm . Since Mm is assumed to be flat, we have (N ⊗R M )m = Nm ⊗Rm Mm → Pm ⊗Rm Mm = (P ⊗R M )m . Applying Proposition 175 we conclude that N ⊗R M → P ⊗R P is injective, and therefore M is flat over R. Corollary 178. Let R be a ring, S ⊂ R a multiplicative subset. If M is a flat R-module, then S −1 M is a flat S −1 R-module. Proof. If M is flat, so is Mp for each prime ideal p of M , but since the primes of S −1 R are a subset of the primes of R, this implies that S −1 M is flat. When a property P of rings or modules is not local, it is often of interest to study also its “localized version”: we say that an R-module M is locally P if for all prime ideals p of R, Mp has property p (and similarly for rings). 7.7.2. Absolute flatness revisited. Lemma 179. A local ring (R, m) is absolutely flat iff it is a field. 33In mantra form: “being isomorphic” is not a local property, but “being an isomorphism” is! 118 PETE L. CLARK Proof. Certainly a field is absolutely flat. The idea of the converse is that by Proposition XX, an absolutely flat ring must have many idempotent elements, whereas a local ring can have no nontrivial idempotents. More precisely, let R be a local ring which is not a field, and let x be a nonzero, nonunit element. Then I = (x) is a proper ideal, and by Proposition X.X we would have R ∼ I ⊕ J . = Proposition 180. Let R be a ring. a) If R is absolutely flat and S ⊂ R is any multiplicative subset, then S −1 R is absolutely flat. b) R is absolutely flat iff for every maximal ideal m of R, Rm is a field. Proof. a) By Exercise X.X, every S −1 R-module is of the form S −1 R ⊗R M for some R-module M . By hypothesis M is flat, so by Corollary 178, so is S −1 M . b) If R is absolutely flat, and m is a maximal ideal of R, then by part a) Rm is absolutely flat. On the other hand it is a local ring, so by Lemma 179 Rm is a field. Conversely, assume that each Rm is a field, and let M be an R-module. Then for all m ∈ MaxSpec R, Mm is a flat Rm -module, so M is a flat R-module. Of course it is natural to ask about whether freeness and projectivity are local properties. We devote the following section to an analysis of this question. 7.8. Local characterization of finitely generated projective modules. 7.8.1. Z-local properties. Let us call a family of {fi }i∈I of elements of R a Z-family if ⟨fi ⟩ = 1. Clearly for every Z-family there is a finite subset J ⊂ I such that {fi }i∈J is also a Z-family. (Later on, this trivial observation will be dressed up in rather fancy attire: this gives the quasi-compactness of the Zariski topology on Spec R.) A property P of rings or modules will be said to be Z-local if it holds over R iff it holds over all Rfi for some Z-family {fi } of R. Proposition 181. Let u : M → N be a homomorphism of R-modules, and let p ∈ Spec R. a) If N is finitely generated and up is surjective, there exists f ∈ R \ p such that uf : Mf → Nf is surjective. b) The surjectivity of u is a Z-local property. c) If M is finitely generated, N is finitely presented and up is an isomorphism, then there exists f ∈ R \ p such that uf : Mf → Nf is an isomorphism. d) If M is finitely generated and N is finitely presented, then the bijectivity of u is a Z-local property. Proof. Write out the exact sequence u 0 → ker u → M → N → coker u → 0. By the flatness of localization, this sequence remains exact upon being tensored with Rf for any f ∈ R or with Rp for any p ∈ R. It follows that passage to the kernel and cokernel commutes with localization. a) We’re assuming 0 = coker(up ) = (coker u)p , i.e., for each x ∈ coker u there exists fx ∈ R \ p such that fx x = 0. Since coker u is a quotient of the finitely generated module N , it is finitely generated, say by x1 , . . . , xn . Then f = fx1 · · · fxn ∈ R \ p COMMUTATIVE ALGEBRA 119 is such that f coker u = 0, so 0 = (coker u)f = coker(uf ) and uf is surjective. b) It is clear that if u is surjective, then for any f ∈ R, uf is surjective. Conversely, let {fi }i∈I be a Z-family such that ufi is surjective for all i. Then for any p ∈ Spec R there exists i ∈ I such that fi ∈ R \ p, so that up is a further localization of ufi and thus the surjectivity of ufi implies that of up . By Proposition 175, u is surjective. c) By part a), there exists f1 ∈ R \ p such that coker uf1 = 0, and thus we have an exact sequence 0 → (ker u)f1 → Mf1 → Nf1 → 0. Since N is finitely presented over R, Nf1 is finitely presented over Rf1 and thus (ker u)f1 is finitely generated. Arguing as in part b), we get f2 ∈ R \ p such that ∼ f1 f2 ker u = 0. Taking f = f1 f2 we get uf : Mf → Nf . d) This is proved analogously to part b) and is left to the reader. Corollary 182. For a finitely presented R-module M , TFAE: (i) There is a Z-family {fi }i∈I of R such that for all i ∈ I , Mfi is a free Rfi -module. (ii) For every prime ideal p of R, Mp is a free Rp -module. (iii) For every maximal ideal m of R, Mm is a free Rm -module. Proof. (i) =⇒ (ii): For each prime ideal p there exists at least one i such that fi ∈ p; equivalently, the multiplicative subset generated by fi is contained in R \ p. / Thus Mp = Mfi ⊗Rfi Rp and since Mfi is free, so is Mp . (ii) =⇒ (i): It is enough to find for each prime ideal p an element fp ∈ R \ p such that Mfp is free: for if so, then {fp }p∈Spec R is a Z-family. Choose x1 , . . . , xn ∈ M whose images in Mp give an Rp -basis, and define u : Rn → M via ei → xi . Then up is an isomorphism, so by Proposition 181c) we may choose fp ∈ R \ p such that ufp is an isomorphism and thus Mfp is free. (ii) ⇐⇒ (iii): this follows from Proposition 175. Exercise 7.14: Let R1 , . . . , Rn be rings and for 1 ≤ i ≤ n, Mi a finitely generated ∏n projective Ri -module. Show that M = i=1 Mi is a finitely generated projective ∏n R = i=1 Ri -module. We can now prove one of the major results of this text. Theorem 183. Let R be a ring and M an R-module. The following are equivalent: (i) M is finitely generated and projective. (ii) M is finitely presented and for all m ∈ MaxSpec R, Mm is a free Rm -module. (iii) For every maximal ideal m of R, there exists f ∈ R \ m such that Mf is a locally free Rf -module of finite rank. (iv) There exists a finite Z-family {f1 , . . . , fn } of R such that ⟨f1 , . . . , fn ⟩ = R and for all i, Mfi is a finitely generated free Rfi -module. Proof. (i) =⇒ (ii): Let M be finitely generated and projective. There exists a finitely generated free module F and a surjection q : F → M . Since M is projective, q splits and Ker(q ) is not just a submodule of F but also a quotient and thus finitely generated. So M is finitely presented. Since projectivity is preserved by base change and any finitely generated projective module over a local ring is free (Theorem 24), for all maximal ideals m of R, Mm is free. (ii) =⇒ (iii): this follows immediately from Corollary 182. (iii) =⇒ (iv): For each m ∈ MaxSpec R, choose fm ∈ R \ m such that Mfm is a finitely generated free Rfm -module. Then {fm }m∈MaxSpec R is a Z-family of R, and 120 PETE L. CLARK as remarked above, every Z-family contains a finite subfamily. ∏n (iv) =⇒ (i): Put S = i=1 Rfi and let f : R → S be the natural map. Step 1: First note that ker f = n ∩ i=1 ker(R → Rfi ) = n ∩ ann(fi ) = ann⟨f1 , . . . , fn ⟩ = ann R = 0, i=1 so f is injective, and thus S is an extension ring of R. Step 2: We claim f : R → S is a faithfully flat extension. Since localizations are flat and direct sums of flat algebras are flat, S/R is a flat extension. So by Theorem 106, it is enough to show that f ∗ : Spec S → Spec R is surjective. But ⨿n Spec S = i=1 Spec Rfi and f ∗ (Spec Rfi ) is the subset of p ∈ Spec R such that fi ∈ p. Since {f1 , . . . , fn } forms a Z-family, no proper ideal can contain all the fi ’s, / and therefore p lies in at least one f ∗ (Spec Rfi ). Step 3: We have ∏faithfully flat ring extension f : R → S and an R-module M such a ∏n n that M ⊗R S = i=1 Mfi is finitely generated and projective as an S = i=1 Rfi module (Exercise X.X). By Theorem 109, M is finitely generated and projective! Corollary 184. Every finitely presented flat R-module is projective. Proof. Let M be a finitely presented, flat R-module. For each maximal ideal m of R, Mm is a finitely presented flat module over the local ring Rm , hence is free by Theorem 58. Therefore by criterion (iii) of Theorem 183, M is projective. Corollary 185. Let M be finitely generated over the Noetherian ring R. TFAE: (i) M is projective. (ii) M is locally free. (iii) M is flat. Exercise 7.15: Prove Corollary 185. Corollary 185 is the last word on finitely generated projective modules over Noetherian rings. In the non-Noetherian case, Corollary 184 leaves a little room for improvement: could it be true that every finitely generated flat module is projective? This is not true in general but it is true for some important classes of non-Noetherian rings, e.g. any connected ring. To see this we need to make a topological study of the rank function on a finitely generated projective module. This is taken up later on in §X.X . Theorem 186. For an R-module M , TFAE: (i) M is finitely generated projective. (ii) For all R-modules B , the natural map Φ : A∨ ⊗R B → HomR (A, B ) induced by (f, b) → (a → f (a)b) is an isomorphism. (iii) The map Φ : A∨ ⊗R A → HomR (A, A) is an isomorphism. Proof. (i) =⇒ (ii): It is enough to show that for all p ∈ Spec R, Φp is an isomorphism. Since M is finitely generated projective, it is finitely presented; moreover Rp is a flat R-module, so by Theorem 101 we have a canonical isomorphism HomR (M, N ) ⊗R Rp = HomRp (Mp , Np ). Also tensor products commute with base change, so it is enough to show Φp : A∨ ⊗Rp Bp → HomRp (Ap , Bp ) p COMMUTATIVE ALGEBRA 121 is an isomorphism. Since A is finitely generated projective, Ap is finitely generated and free. We are thus essentially reduced to a familiar fact from linear algebra, ∼ namely the canonical isomorphism V ∨ ⊗ W → Hom(V, W ) for vector spaces over a field, with V finite-dimensional. We leave the details to the reader as an exercise. (ii) =⇒ (iii): This is immediate. ∑m −1 (iii) ∑m =⇒ (i): Let Φ (1M ) = i=1 fi ⊗ ai . Then we have that for all a ∈ M , a = i=1 fi (a)ai . By the Dual Basis Lemma, M is finitely generated projective. 7.8.2. Infinitely generated locally free modules. Let M be an R-module which is not necessarily finitely generated. Since projectivity is preserved by base change and by Theorem X.X every projective module over a local ring is free, it follows that if M is projective it is locally free. What about the converse? It need not hold: for infinitely generated modules, being locally free can be a much weaker property. Consider: Proposition 187. For a ring R with MaxSpec R = Spec R, TFAE: (i) R is absolutely flat. (ii) Every R-module is locally free. Proof. (i) =⇒ (ii): By Propoition 180, for m ∈ MaxSpec R, Rm is a field, so every Rm -module is free. Since MaxSpec R = Spec R, every R-module is locally free. (ii) =⇒ (i): Again by Proposition 180, if R is not absolutely flat, then there exists some m ∈ MaxSpec R such that Rm is not a field, and thus there exists a nonfree Rm -module Mm . By Exercise X.X, there is an R-module M such that M ⊗R Rm ∼ Rm and thus M is not locally free. = Remark: Later we will see that a ring R is absolutely flat iff every prime ideal of R/ nil(R) is maximal. In particular, for every absolutely flat ring we necessarily have MaxSpec R = Spec R, so Proposition 187 is actually telling us that a ring is absolutely flat iff it is “absolutely locally free”. As we have seen, there are plenty of rings which are absolutely flat but not absolutely projective. For instance an absolutely projective ring is Noetherian, so an infinite product of fields or an infinite Boolean ring will carry locally free, nonprojective modules. 8. Noetherian rings We have already encountered the notion of a Noetherian ring, i.e., a ring in which each ideal is finitely generated; or equivalently, a ring which satisfies the ascending chain condition (ACC) on ideals. Our results so far have given little clue as to the importance of this notion. But in fact, as Emmy Noether showed, consideration of rings satisfying (ACC) is a major unifying force in commutative algebra. In this section we begin to see why this is the case. After giving an introductory examination of chain conditions on rings and modules, we are able to make the key definitions of height of a prime ideal and dimension of a ring, which we will slowly but surely work towards understanding throughout the rest of these 122 PETE L. CLARK notes. Indeed we begin by giving a reasonably complete analysis of the structure theory of Artinian rings, which, as we will show, really is our first order of business in attempting the systematic study of Noetherian rings, since according to the Akizuki-Hopkins theorem the Artinian rings are precisely the Noetherian rings of dimension zero. We are then able to state and prove three of the most important and useful theorems in the entire subject. Whereas the first theorem, the Hilbert basis theorem, gives us a large supply of Noetherian rings, the latter two theorems, Krull’s intersection theorem and Krull’s principal ideal theorem, are basic results about the structure theory of Noetherian rings. 8.1. Chain conditions on posets. Let S be a partially ordered set. S satisfies the Ascending Chain Condition (ACC) if there is no infinite sequence {xi }∞ of elements of S such that xi < xi+1 i=1 for all i ∈ Z+ . One says there is no infinite properly ascending chain. Similarly, S satisfies the Descending Chain Condition (DCC) if there is no infinite sequence {yj }∞ of elements of S such that yj > yj +1 for all j ∈ Z+ . j =1 Of course, at the level of abstract partially ordered sets these two notions are essentially equivalent: a partially ordered set S satisfies (ACC) iff S op – the same underlying set endowed with the opposite ordering x ≤′ y ⇐⇒ y ≤ x – satisfies (DCC). Nevertheless in our applications the two conditions remain quite distinct. Examples: If S is finite it satisfies both ACC and DCC. With the usual orderings, the positive integers Z+ satisfy DCC but not ACC, the negative integers Z− (or equivalently, Z+ with the opposite ordering) satisfy ACC but not DCC, and the integers Z satisfy neither. Exercise 8.1: Let S be a poset. a) Show that S satisfies (ACC) (resp. (DCC)) iff there is no order embedding Z+ → S (resp. Z− → S ). b) Suppose S is totally ordered. Show that S satisfies (DCC) iff it is well-ordered: i.e., every nonempty subset has a minimal element. Note that strictly speaking it would be more accurate to speak of ascending and descending sequences, since generally speaking by a “chain” in a partially ordered set one means any totally ordered subset. Thus both infinite ascending and descending sequences are examples of chains, and the terms “ascending” and “descending” cannot be meaningfully a applied to an arbitrary chain. 8.2. Chain conditions on modules. Let R be a ring, and M a (left) R-module. It makes sense to speak of the (ACC) and (DCC) for R-submodules of M . Indeed, we will call M a Noetherian module if it satisfies (ACC) and an Artinian module if it satisfies (DCC). Exercise 8.2: Show that an R-module M is Noetherian iff every R-submodule M ′ of M is finitely generated. Example: As a Z-module, the integers Z are Noetherian but not Artinian. COMMUTATIVE ALGEBRA 123 Example: As a Z-module, the group of all p-power roots of unity in the complex numbers – in other words, limn→∞ µpn – is Artinian but not Noetherian. Notice that every ring R is naturally an R-module, and the R-submodules of R are precisely the ideals. Thus it makes sense to say whether R is a Noetherian or Artinian R-module, and – thank goodness – this is visibly consistent with the previous terminology. Exercise 8.3: Let M ′ ⊂ M be R-modules, and φ : M → M/M ′ be the quotient map. If N1 and N2 are submodules of M such that N1 ⊂ N2 , N1 ∩ M ′ = N2 ∩ M ′ and φ(N1 ) = φ(N2 ), show that N1 = N2 . Theorem 188. Let 0 → M ′ → M → M ′′ → 0 be a short exact sequence of R-modules. Then M is Noetherian (resp. Artinian) iff both M ′ and M ′′ are Noetherian (resp. Artinian). Proof. We do the Noetherian case, leaving the similar Artinian case as an exercise for the reader. First, since an infinite ascending chain in a submodule or quotient module of M gives rise to an infinite ascending chain in M , if M is Noetherian, both M ′ and M ′′ are. Conversely, suppose N1 N2 . . . is an infinite ascending chain of submodules of M . Consider the chain (Ni + M ′ )/M ′ in M ′′ = M/M ′ . By hypothesis, this chain eventually stabilizes, i.e., for sufficiently large i and j , Ni + M ′ = Nj + M ′ . Similarly, by intersecting with M ′ we get that for sufficiently large i and j Ni ∩ M ′ = Nj ∩ M ′ . Applying Exercise X.X we conclude Ni = Nj for all sufficiently large i, j . A ring R is Noetherian if R is a Noetherian R-module. A ring R is Artinian if R is an Artinian R-module. Exercise 8.3.5: Let R be a ring. a) Show that R is Noetherian iff every finitely generated R-module is Noetherian. b) Show that R is Artinian iff every finitely generated R-module is Artinian. c) Exhibit a ring R which is Noetherian but not Artinian. d) Can you find a ring R which is Artinian but not Noetherian?34 8.3. Semisimple modules and rings. In this section we allow not necessarily commutative rings R. By a “module over R” we mean a left R-module unless otherwise indicated. A module M is simple if it is nonzero and has no proper, nonzero submodules. This definition is of course made in analogy to that of a simple group, namely a nontrivial group possessing no nontrivial proper normal subgroups. And indeed many of the results in this and subsequent sections were first proved in the context of groups. It is even possible to work in a single context that simultaneously generalizes the case of groups and modules (over a not necessarily commutative ring), 34More on this later! 124 PETE L. CLARK the key concept being that of groups with operators. For more on this perspective we invite the reader to consult any sufficiently thick all-purpose graduate level algebra text, the gold standard here being [J1], [J2]. Exercise 8.4 (Schur’s Lemma): Let M be a simple R-module. Show that EndR (M ) is a division ring. Theorem 189. For an R-module M , TFAE: (i) M is a direct sum of simple submodules. (ii) Every submodule of M is a direct summand. (iii) M is a sum of simple submodules. A modules satisfying these equivalent conditions is called semisimple. ⊕ Proof. (i) =⇒ (ii): Suppose M = i∈I Si , with each Si a simple submodule. For ⊕ each J ⊂ I , put MJ = i∈J Si . Now let N be an R-submodule of M . An easy Zorn’s Lemma argument gives us a maximal subset J ⊂ I such that N ∩ MJ = 0. For i ∈ J we have (MJ ⊕ Si ) ∩ N ̸= 0, so choose 0 ̸= x = y + z , x ∈ N , y ∈ MJ , / z ∈ Si . Then z = x − y ∈ (Mj + N ) ∩ Si , and if z = 0, then x = y ∈ N ∩ Mj = 0, contradiction. So (MJ ⊕ N ) ∩ Si ̸= 0. Since Si is simple, this forces Si ⊂ MJ ⊕ N . It follows that M = MJ ⊕ N . (ii) =⇒ (i): First observe that the hypothesis on M necessarily passes to all submodules of M . Next we claim that every nonzero submodule C ⊂ M contains a simple module. proof of claim: Choose 0 ̸= c ∈ C , and let D be a submodule of C which is maximal with respect to not containing c. By the observation of the previous paragraph, we may write C = D ⊕ E . Then E is simple. Indeed, suppose not and let 0 F E . Then E = F ⊕ G so C = D ⊕ F ⊕ G. If both D ⊕ F and D ⊕ G contained c, then c ∈ (D ⊕ F ) ∩ (D ⊕ G) = D, contradiction. So either D ⊕ F or D ⊕ G is a strictly larger submodule of C than D which does not contain c, contradiction. So E is simple, establishing our claim. Now let N ⊂ M be maximal with respect to being a direct sum of simple submodules, and write M = N ⊕ C . If C ̸= 0, then by the claim C contains a nonzero simple submodule, contradicting the maximality of N . Thus C = 0 and M is a direct sum of simple submodules. (i) =⇒ (iii) is immediate. (iii) =⇒ (i): as above, by Zorn’s Lemma there exists a submodule N of M which is maximal with respect to being a direct sum of simple submodules. We must show N = M . If not, since M is assumed to be generated by its simple submodules, there exists a simple submodule S ⊂ M which is not contained in N . But since S is simple, it follows that S ∩ N = 0 and thus N ⊕ S is a strictly larger direct sum of simple submodules: contradiction. Corollary 190. An R-module M has a unique maximal semisimple submodule, called the socle of M and written Soc M . Thus M is semisimple iff M = Soc M . Exercise 8.5: Prove Corollary 190. Exercise 8.6: Let N ∈ Z+ . Compute the socle of the Z-module Z/N Z. Show in particular that Z/N Z is semisimple iff N is squarefree. COMMUTATIVE ALGEBRA 125 A not necessarily commutative ring R is left semisimple if R is semisimple as a left R-module. Theorem 191. For a nonzero not necessarily commutative ring R, TFAE: (i) R is left semisimple. (ii) Every left ideal of R is a direct summand. (iii) Every left ideal of R is an injective module. (iv) All left R-modules are semisimple. (v) All short exact sequences of left R-modules split. (vi) All left R-modules are projective. (vii) All left R-modules are injective. Proof. We will show (i) ⇐⇒ (ii), (iv) ⇐⇒ (v) ⇐⇒ (vi) ⇐⇒ (vii) and (ii) =⇒ (vii) =⇒ (iii) =⇒ (ii), which suffices. (i) =⇒ (ii) follows immediately from Theorem 189. (iv) ⇐⇒ (v) follows immediately from Theorem 189. (v) ⇐⇒ (vi) and (v) ⇐⇒ (vii) are immediate from the definitions of projective and injective modules. (ii) =⇒ (vii): Let I be a left ideal of R and f : I → M an R-module map. By π1 hypothesis, there exists J such that I ⊕ J = R, so f extends to F : R = I ⊕ J → I → M . By Baer’s Criterion, M is injective. (vii) =⇒ (iii) is immediate. (iii) =⇒ (ii) is immediate from the definition of injective modules. Lemma 192. Let R be a ring and {Mj }j ∈J be an indexed family of nonzero Rmodules. The following are equivalent: (i) I is finite and each Mj is finitely generated. ⊕ (ii) M = j ∈J Mj is finitely generated. Proof. (i) =⇒ (ii) is left to the reader as an easy exercise. (ii) =⇒ (i): Each Mj is isomorphic to a quotient of M , so if M is finitely generated, so is Mj . Now let X = {x1 , . . . , xn } be a finite generating set for M , and for each ∑ 1 ≤ 1 ≤ n, let xij be the j -component of xi , so xi = j ∈J xij . This sum is of course finite, and therefore the set J ′ ⊂ J of indices j such that xij ̸= 0 for some ⊕ 1 ≤ i ≤ n is finite. It follows that ⟨X ⟩ ⊂ j ∈J ′ Mj M , contradiction. Lemma 193. Let R1 , . . . , Rn be finitely many not necessarily commutative rings, ∏n and put R = i=1 Ri . Then R is semisimple iff Ri is semisimple for all 1 ≤ i ≤ n. Exercise: Prove Lemma 193. We now quote the following basic result from noncommutative algebra. Theorem 194. (Wedderburn-Artin) For a ring R, TFAE: (i) R is semisimple as a left R-module (left semisimple). (ii) R is semisimple as a right R-module (right semisimple). (iii) There are N, n1 , . . . , nN ∈ Z+ and division rings D1 , . . . , DN such that R∼ = N ∏ i=1 Mni (Di ). 126 PETE L. CLARK Combining Theorems 191 and 194 gives us a tremendous amount of information. First of all, a ring is left semisimple iff it is right semisimple, so we may as well speak of semisimple rings. A ring is semisimple iff it is absolutely projective iff it is absolutely injective. Coming back to the commutative case, the Wedderburn-Artin theorem tells us that the class of semisimple / absolutely projective / absolutely injective rings is extremely restricted. Corollary 195. A commutative ring is semisimple iff it is a finite product of fields. However it is significantly easier to give a proof of Wedderburn-Artin in the commutative case, so we will give a direct proof of Corollary 195 Proof. Step -1: Officially speaking the theorem holds for the zero ring because it is an empty product of fields. In any event, we may and shall assume henceforth that our semisimple ring is nonzero. Step 0: A field is a semisimple ring: e.g. every module over a field is free, hence projective. By Lemma 193, a finite direct product of fields is therefore semisimple. ⊕ Step 1: Let R be a semisimple ring, and let R = i∈I Mi be a direct sum decomposition into simple R-modules. R is a finitely generated R-module, by Lemma 192 I is finite, and we may identify it with {1, . . . , n} for some n ∈ Z+ : R = M1 ⊕ . . . ⊕ Mn . Step 2: We may uniquely write 1 = e1 + . . . + en with ei ∈ Mi . Then for all i ̸= j , ei ej = 0, and this together with the identity 1 · 1 = 1 implies that e2 = ei for all i i. As usual for idempotent decompositions, this expresses R as a direct product of the subrings Ri = Mi = ei R. Moreover, since Mi is a simple R-module, Ri has no proper nonzero ideals, and thus it is a field, say ki . Exercise 8.7: Exhibit an absolutely flat commutative ring which is not semisimple. 8.4. Normal Series. If M is an R-module a normal series is a finite ascending chain of R-submodules 0 = M0 M1 . . . Mn = M . We say that n is the length of the series. (The terminology is borrowed from group theory, in which one wants a finite ascending chain of subgroups with each normal in the next. Of course there is no notion of “normal submodule”, but we keep the group-theoretic terminology.) There is an evident partial ordering on the set of normal series of a fixed R-module ′ ′ M : one normal series {Mi }n is less than another normal series {Mj }n=0 if for all i=0 j ′ 1 ≤ i ≤ n, Mi is equal to Mj for some (necessarily unique) j . Rather than saying ′ ′ that {Mi } ≤ {Mj }, it is traditional to say that the larger series {Mj } refines the smaller series {Mi }. Given any normal series {Mi } we may form the associated factor sequence ′ ′ M1 /M0 = M1 , M2 /M1 , . . . , Mn /Mn−1 = M/Mn−1 . Two normal series {Mi }n , {Mj }n=0 i=0 j are equivalent if n = n′ and there is a permutation σ of {1, . . . , n} such that for all ′ ′ 1 ≤ i ≤ n, the factors Mi /Mi−1 and Mσ(i) /Mσ(i)−1 are isomorphic. In other words, if we think of the factor sequence of a normal series as a multiset of isomorphism classes of modules, then two normal series are equivalent if the associated multisets of factors are equal. COMMUTATIVE ALGEBRA 127 Exercise 8.8: Show that refinement descends to a partial ordering on equivalence classes of normal series of a fixed R-module M . The following theorem is the basic result in this area. Theorem 196. (Schreier Refinement) For any R-module M , the partially ordered set of equivalence classes of normal series of submodules of M is directed: that is, any two normal series admit equivalent refinements. Proof. For a proof in a context which simultaneously generalizes that of modules and groups, see e.g. [J2, p. 106]. For an R-module M , a composition series is a maximal element in the poset of normal series: that is, a composition series which admits no proper refinement. Exercise 8.9: Show that a normal series {Mi }n for an R-module M is a comi=0 position series if and only if for all 1 ≤ i ≤ n, the factor module Mi /Mi−1 is simple. o Theorem 197. (Jordan-H¨lder) Let M be an R-module. Then any two composition series for M are equivalent: up to a permutation, their associated factor series are term-by-term isomorphic. Proof. This is an immediate consequence of Schreier Refinement: any two normal series admit equivalent refinements, but no composition series admits a proper refinement, so any two composition series must already be equivalent! Remark: The order-theoretic core of the proof is simply that a directed set admits at most one maximal element. Thus for a module M which admits a composition series, we may define the length ℓ(M ) of M to be the length of any composition series. One also speaks of the Jordan-H¨lder factors of M or the composition factors of M, i.e., the unique o multiset of isomorphism classes of simple R-modules which must appear as the successive quotients of any composition series for M . If a module does not admit a composition series, we say that it has infinite length. And now a basic question: which R-modules admit a composition series? Exercise 8.10: a) Show that any finite35 module admits a composition series. b) Show that if a module M admits a composition series, it is finitely generated. c) Show that a Z-module M admits a composition series iff it is finite. d) Let k be a field. Show that a k -module admits a composition series iff it is finitely generated (i.e., iff it is finite-dimensional). Exercise 8.11: An R-module M admits a composition series iff there exists L ∈ Z+ such that every normal series in M has length at most L. 35Recall that in these notes by a “finite module” we mean a module whose underlying set is finite! 128 PETE L. CLARK Theorem 198. For an R-module M , TFAE: (i) M is both Noetherian and Artinian. (ii) M admits a composition series. Proof. Assume (i). Since M satisfies (DCC), there must exist a minimal nonzero submodule, say M1 . If M1 is a maximal proper submodule, we have a composition series. Otherwise among all proper R-submodules strictly containing M1 , by (DCC) we can choose a minimal one M2 . We continue in this way: since M also satisfies (ACC) the process must eventually terminate, yielding a composition series. (ii) =⇒ (i): This follows easily from Exercise 8.11. Exercise 8.12: Exercise 8.11 (and hence also our proof of Theorem 198) makes use of Schreier Refinement. Give a proof that (ii) =⇒ (i) in Theorem 198 which is independent of Schreier Refinement. (Suggestion: try induction on the length of a composition series.) Proposition 199. Let 0 → M ′ → M → M ′′ → 0 be a short exact sequence of R-modules. Then: a) M admits a composition series iff both M ′ and M ′′ admit composition series. b) If M admits a composition series, then ℓ(M ) = ℓ(M ′ ) + ℓ(M ′′ ). Exercise 8.13: Prove Proposition 199. Remark: Although it will not play a prominent role in our course, the length of an R-module M is an extremely important invariant, especially in algebraic geometry: it is is used, among other things, to keep track of intersection multiplicities and to quantitatively measure the degree of singularity of a point. 8.5. The Krull-Schmidt Theorem. The material in this section follows [J2, §3.4] very closely. In particular, very exceptionally for us – but as in loc. cit. – in this section we work with left modules over a possibly noncommutative ring R. The reason: not only does the desired result carry over verbatim to the noncommutative case (this is not in itself a good enough reason, as the same holds for a positive proportion of the results in these notes) but the proof requires us to consider noncommutative rings! A module M is decomposable if there are nonzero submodules M1 , M2 ⊂ M such that M = M1 ⊕ M2 ; otherwise M is indecomposable. Theorem 200. (Krull-Schmidt) Let M be an R-module of finite length. Then: ⊕m a) There are indecomposable submodules M1 , . . . , Mm such that M = i=1 Mi . ⊕n b) If there are indecomposable submodules N1 , . . . , Nn such that M = i=1 Ni , then m = n and there exists a bijection σ of {1, . . . , n} such that for all i, Mi ∼ Nσ(i) . = The Proof of Theorem 200a) is easy, and we give it now. If M is a finite length module and we write M = M1 ⊕ M2 then 0 < ℓ(M1 ), ℓ(M2 ) < ℓ(M ). Thus an evident induction argument shows that any sequence of moves, each one of which splits a direct summand of M into two nontrivial direct subsummands of M , must terminate after finitely many steps, leaving us with a decomposition of M into a finite direct sum of indecomposable submodules. COMMUTATIVE ALGEBRA 129 As one might suspect, the second part of Theorem 200 concerning the uniqueness of the indecomposable decomposition is more subtle. Indeed, before giving the proof we need some preparatory considerations on endomorphism rings of modules. Proposition 201. For an R-module M , TFAE: (i) M is decomposable. (ii) The (possibly noncommutative, even if R is commutative) ring EndR (M ) = HomR (M, M ) has a nontrivial idempotent, i.e., an element e ̸= 0, 1 with e2 = e. Exercise 8.14: Prove Proposition 201. A (not necessarily commutative) ring R is local if the set of nonunits R \ R× forms a two-sided ideal of R. Exercise 8.15: Let R be a local, not necessarily commutative ring. a) Show that R ̸= 0. b) Show that R has no nontrivial idempotents. An R-module M is strongly indecomposable if EndR (M ) is local. Thus it follows from Proposition 201 and Exercise 8.15 that a strongly indecomposable module is indecomposable. Example: The Z-module Z is indecomposable: any two nonzero submodules (a) and (b) have a nontrivial intersection (ab). On the other hand EndZ (Z) = Z is not a local ring, so Z is not strongly indecomposable. Thus “strongly indecomposable” is, in general, a stronger concept than merely “indecomposable”. Notice though that the Krull-Schmidt theorem applies only to finite length modules – equivalently to modules which are both Noetherian and Artinian – and Z is not an Artinian Z-module. In fact, it shall turn out that any finite length indecomposable module is strongly indecomposable, and this will be a major step towards the proof of the Krull-Schmidt Theorem. But we are not quite ready to prove this either! First some Fitting theory. For an R-module M and f ∈ EndR (M ), we put f ∞ (M ) = ∞ ∩ f n (M ). n=1 ∞ The set f (M ) is the intersection of a descending chain M ⊃ f (M ) ⊃ f 2 (M ) ⊃ . . . ⊃ f n (M ) ⊃ . . . of submodules of M , and is thus an f -stable submodule of M . The restriction of f to f ∞ (M ) is surjective. Moreover, if M is an Artinian module, there exists s ∈ Z+ such that f s (M ) = f s+1 (M ) = . . .. Exercise 8.16: Find a commutative ring R, an R-module M and f ∈ EndR (M ) such that for no n ∈ Z+ is the submodule f n (M ) f -stable. 130 PETE L. CLARK Similarly, for M and f as above, we put f−∞ (0) = ∞ ∪ ker f n . n=1 Here each ker f n is an f -stable submodule of M on which f is nilpotent. The set f−∞ (0) is the union of an ascending chain of submodules 0 ⊂ ker f ⊂ ker f 2 ⊂ . . . ⊂ ker f n ⊂ . . . of M and is thus an f -stable submodule of M on which f acts as a nil endomorphism: i.e., every element of M is killed by some power of f . Moreover, if M is a Noetherian module, there exists t ∈ Z+ such that ker f t = ker f t+1 = . . . and thus f is a nilpotent endomorphism of f−∞ (0). Exercise 8.17: Find a commutative ring R, an R-module M and f ∈ EndR (M ) such that f is not a nilpotent endomorphism of f−∞ (0). Theorem 202. (Fitting’s Lemma) Let M be a finite length module over the not necessarily commutative ring R, and let f ∈ EndR (M ). a) There exists a Fitting Decomposition (9) M = f ∞ (M ) ⊕ f −∞ (0). b) f |f ∞ (M ) is an isomorphism and f |f−∞ (0) is nilpotent. Proof. Since M has finite length it is both Noetherian and Artinian. Thus there exists r ∈ Z+ such that f r (M ) = f r+1 (M ) = . . . = f ∞ (M ) and ker f r = ker f r+1 = . . . = f−∞ (0). Let x ∈ f ∞ (M ) ∩ f−∞ (0). Then there exists y ∈ M such that x = f r (y ); moreover 0 = f r (x) = f 2r (y ). But f 2r (y ) = 0 implies x = f r (y ) = 0, so f ∞ (M )∩f−∞ (0) = 0. Let x ∈ M . Then f r (x) ∈ f r (M ) = f 2r (M ), so there exists y ∈ M with r f (x) = f 2r (y ) and thus f r (x − f r (y )) = 0. so x = f r (y ) + (x − f r (y )) ∈ f ∞ (M ) + f −∞ (0), completing the proof of part a). As for part b), we saw above that the restriction of f to f ∞ (M ) is surjective. It must also be injective since every element of the kernel lies in f −∞ (0). Thus f |f ∞ (M ) is an isomorphism. Finally, as observed above, since f−∞ (0) = ker f r , f |f−∞ (0) is nilpotent. Lemma 203. Let x and y be nilpotent elements in a not necessarily commutative ring (which is not the zero ring). Then x + y is not a unit of R. Proof. Assume to the contrary that x + y = u ∈ R× . Dividing through by u we reduce to showing that we cannot have two nilpotent elements∑ y ∈ R such that x, ∞ x + y = 1. But if x is nilpotent, the “infinite geometric series” n=0 1 + x + . . . + xn + . . . is in fact finite, hence perfectly legal in our abstract algebraic context, and it is immediate to check that the familiar calculus identity ∞ ∑ (1 − x)( xn ) = 1 n=0 COMMUTATIVE ALGEBRA 131 holds here. Thus y = 1−x is both a unit of R and a nilpotent element, contradiction. Corollary 204. Let M be a finite length indecomposable R-module. Then every f ∈ EndR (M ) is either an automorphism or nilpotent. Moreover M is is strongly indecomposable. Proof. Since M is indecomposable, Fitting’s Lemma implies that for f ∈ EndR (M ) we must have either M = f ∞ (M ) – in which case f is an automorphism – or M = f −∞ (0) – in which case f is nilpotent. We must show that I = EndR (M ) \ EndR (M )× is a two-sided ideal of EndR (M ). Note that I is precisely the set of endomorphisms of M which are not automorphisms, hence every element of I is nilpotent. By Lemma 203, I is a subgroup of (R, +). Moreover, for f ∈ I, g ∈ EndR (M ), since f is neither injective nor surjective, gf is not injective and f g is not surjective, so neither is an automorphism and both lie in I . Lemma 205. Let M be a nonzero R-module and N an indecomposable R-module. Suppose we have homomorphisms f : M → N, g : N → M such that gf is an automorphism of M . Then both f and g are isomorphisms. Proof. Let h = (gf )−1 , l = hg : N → M and e = f l : N → N . Then lf = hgf = 1M and e2 = f lf l = f 1M l = f l = e. Since M is indecomposable, either e = 1 or e = 0, and the latter implies 1M = 12 = lf lf = lef = 0, i.e., M = 0. So M f l = e = 1N , so f is an isomorphism and thus so too is (f (gf )−1 )−1 = g . ⊕m Theorem 206. Let M ∼ N be isomorphic modules, and let M = = i=1 Mi and ⊕n N = i=1 Ni′ with each Mi strongly indecomposable and each Ni indecomposable. Then m = n and there is a bijection σ of {1, . . . , m} such that for all i, Mi ∼ Nσ(i) . = Proof. By induction on m: m = 1 is clear. Suppose the result holds for all direct sums of fewer than m strongly indecomposable submodules. Step 1: Let e1 , . . . , em ∈ EndR (M ) and f1 , . . . , fn ∈ EndR (N ) be the idempotent elements corresponding to the given direct sum decompositions (i.e., projection ∼ onto the corresponding factor). Let g : M → N , and put hj := fj ge1 ∈ HomR (M, N ), kj = e1 g −1 fj ∈ HomR (N, M ), 1 ≤ j ≤ n. Then n ∑ j =1 kj hj = ∑ e1 g −1 fj ge1 = e1 g −1 j ∑ fj ge1 = e1 g −1 1N ge1 = e1 . j The restrictions of e1 and kj hj to M1 stabilize M1 so may be regarded as endomorphisms of M1 , say e′ and (kj hj )′ , and we have 1 n ∑ (kj hj )′ = e′ = 1M1 . 1 j =1 By assumption EndR M1 is local, so for at least one j , (kj hj )′ is a unit, i.e., an automorphism of M1 . By reordering the Nj ’s we may assume that j = 1, so (k1 h1 )′ ∈ AutR M1 . We may regard the restriction h′ of h1 to M1 as a homomor1 ′ phism from M1 to N1 and similarly the restriction k1 of k1 to N1 as a homomor′′ ′ phism from N1 to M1 , and then k1 h1 = (k1 h1 ) is an automorphism. By Lemma 132 PETE L. CLARK ∼ ∼ ′ 205, h′ = (f1 ge′ ) : M1 → N1 and k1 = (e1 g −1 f1 )′ : N1 → M1 . 1 1 Step 2: We claim that m ⊕ Mi . (10) M = g −1 (N1 ) ⊕ i=2 ⊕m To see this, let x ∈ g N1 ∩ ( i=2 Mi ), so x = g −1 y for some y ∈ N1 . Because ⊕m x ∈ i=2 Mi , e1 x = 0. Thus −1 ′ 0 = e1 x = e1 g −1 y = e1 g −1 f1 y = k1 y = k1 y. ′ Since k1 is an isomorphism, y = 0 and thus x = 0, so the sum in (10) is direct. ⊕m Now put M ′ = g −1 (N1 ) ⊕ i=2 Mi , so we wish to show M ′ = M . Let x ∈ g −1 N1 . Then x, e2 x, . . . , em x ∈ M ′ , so e1 x = (1 − e2 − . . . − em )x ∈ M ′ . So ′ M ′ ⊃ e1 g −1 N1 = e1 g −1 f1 N1 = k1 N1 = k1 N1 = M1 ⊕m and thus M ′ ⊃ i=1 Mi = M . ∼ Step 3: The isomorphism g : M → N carries g −1 N1 onto N1 hence induces an sim isomorphism g−MN1 → N/N1 . Using Step 2, we have 1 n ⊕ j =2 N ∼ M ∼⊕ Mi . = −1 = N1 g N1 i=2 m Ni = We are done by induction. Exercise 8.18: Please confirm that we have proved the Krull-Schmidt Theorem! Exercise 8.19: Let M and N be R-modules such that M × M ∼ N × N . = a) If M and N are both of finite length, show that M ∼ N . = b) Must we have M ∼ N in general? = Remark: Part b) is far from easy! If you give up, see [Cor64]. Exercise 8.20: a) Let R be a PID and M an R-module. a) Show that M has finite length iff it is a finitely generated torsion module. b) Show that a finitely generated torsion module is indecomposable iff it is isomorphic to R/(pa ) for some prime element p of R and some a ∈ Z+ . c) Did you use the structure theorem for finitely generated modules over a PID to prove parts a) and b)? If so, try to prove these results without it. d) Take as given parts a) and b) of this exercise, and use the Krull-Schmidt Theorem to deduce the structure theorem for finitely generated modules over a PID. Remark: Later we will use these ideas to give an independent proof of the structure theorem for finitely generated modules over a PID, using one extra idea: reduction to the case of a local PID, in which case there is only one nonzero prime ideal and the module theory becomes especially simple. 8.6. Some important terminology. All we aspire to do in this section is to introduce some terminology, but it is so important that we have isolated it for future reference. COMMUTATIVE ALGEBRA 133 Let R be a ring and p a prime ideal of R. The height of p is the supremum of all lengths of finite chains of prime ideals of the form p0 p1 ... pn = p (the length of the indicated chain being n; i.e., it is the number of ’s appearing, which is one less than the number of elements). Thus the height is either a non-negative integer or ∞; the latter transpires iff there exist arbitrarily long finite chains of prime ideals descending from p (and of course, this need not imply the existence of an infinite chain of prime ideals descending from p). A prime ideal of height 0 is called a minimal prime. In an integral domain R, the unique minimal prime is (0), so the concept is of interest only for rings which are not domains. If I is a proper ideal of R, we also speak of a minimal prime over I , which means a prime p ⊃ I such that there is no prime ideal q with I q p. Note that p is a minimal prime over I iff p is a minimal prime in the quotient ring R/I . This remark simultaneously explains the terminology “minimal over” and gives a hint why it is useful to study minimal prime ideals even if one is ultimately most interested in integral domains. The dimension of a ring R is the supremum of all the heights of its prime ideals. The full proper name here is Krull dimension of R, which is of course useful when one has other notions of dimension at hand. Such things certainly do exist but will not be considered here. Moreover, as will shortly become apparent, the need to include Krull’s name here so as to ensure that he gets proper recognition for his seminal work in this area is less than pressing. Therefore we use the full name “Krull dimension” only rarely as a sort of rhetorical flourish. One also often speaks of the codimension of a prime ideal p of R, which is the dimension of R minus the height of p. This is especially natural in applications to algebraic geometry, of which the present notes allude to only in passing. Note that this is not necessarily equal to the Krull dimension of R/p – or what is the same as that, the maximal length of a finite chain of prime ideals ascending from p – although in reasonable applications, and especially in geometry, one is certainly entitled to hope (and often, to prove) that this is the case. Remark: All of these definitions would make perfect sense for arbitrary partially ordered sets and their elements, but the terminology is not completely consistent with order theory. Namely, the height of an element in an arbitrary poset is defined as the supremum of lengths of chains descending from that element, but the order theorists would cringe to hear the supremum of all heights of elements called the “dimension” of the poset. They would call that quantity the height of the poset, and would reserve dimension for any of several more interesting invariants. (Roughly, the idea is that a chain of any finite length is one-dimensional, whereas a product of d chains should have dimension d.) 8.7. Introducing Noetherian rings. The following is arguably the most important single definition in all of ring theory. A ring R is said to be Noetherian if the poset I (R) of all ideals of R satisfies the ascending chain condition. 134 PETE L. CLARK Exercise 8.21: a) Show that a finite ring is Noetherian. b) For each cardinal κ ≥ ℵ0 , exhibit a ring of cardinality κ which is not Noetherian. c) A ring R is said to have finite quotients if for any 0 ̸= I in R, R/I is finite. Show that a ring with finite quotients is Noetherian. d) Deduce that Z is Noetherian. Theorem 207. A finitely generated module over a Noetherian ring is Noetherian. Proof. If M is a finitely generated module over R, then we may represent it as Rn /K for some submodule K of Rn . An immediate corollary of the preceding theorem is that finite direct sums of Noetherian modules are Noetherian, and by assumption R itself is a Noetherian R-module, hence so is Rn and hence (again by Theorem XX) so is the quotient Rn /K = M . Thus so long as we restrict to Noetherian rings, submodules of finitely generated modules remain finitely generated. This is extremely useful even in the case of R = Z: a subgroup of a finitely generated abelian group remains finitely generated. Needless(?) to say, this does not hold for all nonabelian groups, e.g. not for a finitely generated free group of rank greater than 1. Theorem 208. (Characterization of Noetherian rings) For a ring R, TFAE: (i) Every nonempty set of ideals of R has a maximal element. (ii) There are no infinite ascending chains I1 I2 ... In ... of ideals of R. (iii) Every ideal of R is finitely generated. (iv) Every prime ideal of R is finitely generated. Proof. Exercise X.X displays the equivalence of (i) and (ii). Assuming (ii), let I be an ideal of R. If I cannot be finitely generated, this means that we can choose a sequence of elements {xi } of I such that 0 ⟨x1 ⟩ ⟨x1 , x2 ⟩ ..., contradicting (ACC). Assuming (iii), considering an infinite ascending chain of ∪ ideals as in (ii) above, put J = i Ii . Then J is finitely generated, say by elements x1 ∈ Ii1 , . . . , xr ∈ Iir . But then taking i to be the largest of the indices i1 , . . . , ir we have ⟨x1 , . . . , xn ⟩ ⊂ Ii ⊂ J = ⟨x1 , . . . , xr ⟩, so J = Ii = Ii+1 = . . . and the chain stabilizes at I . Trivially (iii) =⇒ (iv). That (iv) implies (iii) is the most difficult implication, and is another instance of the meta-principle that ideals which are maximal with respect to some property tend to be prime. Indeed, this will be a consequence of the following result. Lemma 209. Let R be a ring admitting at least one non-finitely generated ideal. a) Then there exists an ideal which is maximal among all ideals which are not finitely generated. b) Any ideal I which is maximal among all ideals in R which are not finitely generated is necessarily prime. COMMUTATIVE ALGEBRA 135 Proof. For part a) we need only observe that the union of a chain of non-finitely generated ideals remains non-finitely generated for similar reasons to the above: a finite generating set for the union would be a finite generating set for some sufficiently large element of the chain. So suppose I is an ideal which is maximal with respect to not being finitely generated, and let x, y ∈ R be such that xy ∈ I . Assume for a contradiction that neither x nor y is in I . Then the ideal I ′ = ⟨I, a⟩ is strictly larger and thus finitely generated: say I ′ = ⟨i1 + a1 x, . . . , in + an x⟩. Now let J be the set of all b ∈ R with bx ∈ I . Then J contains both I and y , so is strictly larger than I and thus finitely generated. We claim I = ⟨i1 , . . . , in , Jx⟩. Indeed, take any z ∈ I . Then z ∈ I ′ , so x = w1 (i1 + a1 x) + . . . + wn (in + an x). Now u1 x1 + . . . + un xn ∈ J , so z ∈ ⟨i1 , . . . , in , Ja⟩. Thus the claim is established, and, since J is finitely generated, so is Ja and thus I , contradiction. Exercise 8.22: Suppose a ring R satisfies the ascending chain condition on prime ideals. Must R be Noetherian? Proposition 210. Let R be a Noetherian ring. a) If I is any ideal of R, the quotient R/I is Noetherian. b) If S ⊂ R is any multiplicative subset, the localization S −1 R is Noetherian. Proof. Any ideal of R/I is of the form J/I for some ideal J ⊃ I of R. By assumption J is finitely generated, hence J/I is finitely generated, so R/I is Noetherian. A similar argument holds for the localization; details are left to the reader. ∏ Exercise 8.23: Let k be a field, let S be an infinite set, and put R = s∈S k , i.e., the infinite product of #S copies of k . Show that R is not Noetherian, but the localization Rp at each prime ideal is Noetherian. Thus Noetherianity is a localizable property but not a local property. In each of the next three subsections we will state and prove an extremely important and general theorem about Noetherian rings. 8.8. The Bass-Papp Theorem. We now present a beautiful characterization of Noetherian rings in terms of properties of injective modules, due independently to Z. Papp [Pa59] and H. Bass [Bas59]. Theorem 211. (Bass-Papp Theorem) For a ring R, TFAE: (i) A direct limit of injective modules is injective. (ii) A direct sum of injective modules is injective. (iii) A countable direct sum of injective modules is injective. (iv) R is Noetherian. Proof. (i) =⇒ (ii): A direct sum is a kind of direct limit. (ii) =⇒ (iii) is immediate. 136 PETE L. CLARK (iii) =⇒ (iv): Let I1 ⊂ I2 ⊂ . . . ⊂ In ⊂ . . . be an infinite ascending chain of ideals ∪ of R, and let I = n In . We define E= ∞ ⊕ E (R/In ). n=1 For n ∈ Z+ , let fn : I → E (R/In ) be the composite map I → R → R/In → ∏ ∏∞ E (R/In ). There is then a unique map f : I → n=1 E (R/In ). But indeed, for each ∏ fixed x ∈ I , x lies in In for sufficiently large n and thus fn (x) = 0. It follows that f actually lands in the direct sum, and we have thus defined a map f : I → E. By hypothesis, E is a countable direct sum of injective modules and therefore injective, so f extends to an R-module map with domain all of R and is thus of the form f (x) = xf (1) = xe for some fixed e ∈ E . Let N be sufficiently large so that for n ≥ N , the nth component en of e is zero. Then for all x ∈ I , 0 = xen = fn (x) = x + In ∈ R/In , and thus x ∈ In . That is, for all n ≥ N , In = I . (iv) =⇒ (i): let {Eα } be a directed system of injective modules with direct limit E . For α ≤ β we denote the transition map from Eα to Eβ by ιαβ and the natural map from Eα to E by ια . We will show E is injective by Baer’s Criterion (Theorem 29), so let I be any ideal of R and consider an R-module map f : I → E . Since R is Noetherian, I is finitely generated, and it follows that there exists an index α such that f (I ) ⊂ ια (Eα ). Let M be a finitely generated submodule of Eα such that f (I ) ⊂ ια (M ). Consider the short exact sequence ι α 0 → K → M → f (I ) → 0. Since M is finitely generated and R is Noetherian, K is finitely generated. Moreover K maps to 0 in the direct limit, so there exists β ≥ α such that ιαβ K = 0. Let M ′ = ιαβ M , so by construction ∼ ιβ : M ′ → f (I ). Taking g = ιβ |−1′ ◦ f we get a map g : I → Eβ such that f = ιβ ◦ g . Since Eβ is M injective, g extends to a map G : R → Eβ and thus F = ιβ ◦ G extends f to R. 8.9. Artinian rings: structure theory. A ring R which satisfies the descending chain condition (DCC) on ideals is called Artinian (or sometimes, “an Artin ring”). Exercise 8.24 a) Show that a ring with only finitely many ideals is Artinian. b) Show that the ring of integers Z is not Artinian. c) Show that a quotient of an Artinian ring is Artinian. d) Show that a localization of an Artinian ring is Artinian. Obviously any finite ring has only finitely many ideals and is Artinian. It is not difficult to give examples of infinite rings with finitely many ideals. For instance, let k be a field and let 0 ̸= f ∈ k [t]. Then R = k [t]/(f ) has only finitely many COMMUTATIVE ALGEBRA 137 a a ideals. Indeed, if we factor f = f1 1 · · · fr r into irreducible factors, then the Chinese Remainder Theorem gives k [t]/(f ) ∼ k [t]/(f a1 ) × . . . × k [t]/(f ar ). = r 1 Each factor ring are precisely k [t]/(fiai ) is a local ring with maximal ideal (f1 ), and the ideals (0) (fi )ai −1 ... fi . Since every ideal in a product is a direct sum of ideals of the factors, there are then ∏r precisely i=1 (ai + 1) ideals of R. A bit of reflection reveals that – notwithstanding their very similar definitions – requiring (DCC) on ideals of a ring is considerably more restrictive than the (ACC) condition. For instance: Proposition 212. A domain R is Artinian iff it is a field. Proof. Obviously a field satisfies (DCC) on ideals. Conversely, if R is a domain and not a field, there exists a nonzero nonunit element a, and then we have (a) (a2 ) (a3 ) . . .. Indeed, if (ak ) = (al ), suppose k ≤ l and write l = k + n, and then we have uak = ak an for some u ∈ A× , and then by cancellation we get an = u, so an is unit and thus a is a unit, contradiction. The result collects several simple but important properties of Artinian rings. Theorem 213. Let R be an Artinian ring. a) R has dimension zero: prime ideals are maximal. b) Therefore the Jacobson radical of R coincides with its nilradical. c) R has only∩ finitely many maximal ideals, say m1 , . . . , mn . n d) Let N = i=1 mi be the nilradical. Then it is a nilpotent ideal: there exists + k ∈ Z such that N k = 0. Proof. a) If p is a prime ideal of A, then A/p is an Artinian domain, which by Proposition XX is a field, so p is maximal. b) By definition, the Jacobson radical is the intersection of all maximal ideals and the nilradical is the intersection of all prime ideals. Thus the result is immediate from paqrt a). c) Suppose mi is an infinite sequence of maximal ideals. Then R m1 m1 ∩ m2 · · · is an infinite descending chain. Indeed, equality at any step would mean mN +1 ⊃ ∏N ∩N i=1 mi , and then since mN +1 is prime it contains mi for some 1 ≤ i ≤ i=1 mi = N , contradiction. d) Applying DCC on the powers of N , it must be the case that there exists some k with N k = N k+n for all n ∈ Z+ . Put I = N k . Suppose I ̸= 0, and let Σ be the set of ideals J such that IJ ̸= 0. Evidently Σ ̸= ∅, for I ∈ Σ. By DCC we are entitled to a minimal element J of σ . There exists 0 ̸= x ∈ J such that xI ̸= 0. For such an x, we have (x) ∈ Σ and by minimality we must have J = (x). But (xI )I = xI ̸= 0, so xI ⊂ (x) and thus xI = (x) by minimality. So there exists y ∈ I with xy = x and thus we have x = xy = xy 2 = . . . = xy k = . . . . 138 PETE L. CLARK But y ∈ I ⊂ N , hence y is nilpotent and the above equations give x = 0, a contradiction. Lemma 214. Suppose that in a ring R there exists a ∏ finite sequence m1 , . . . , mn of (not necessarily distinct) maximal ideals such that 0 = i mi . Then R is Noetherian iff it is Artinian. Proof. Consider the chain of ideals R ⊃ m1 ⊃ m1 m2 ⊃ . . . ⊃ ∏ mi = 0. i Each quotient Qi := m1 · · · mi−1 /m1 · · · mi is a vector space over R/mi . Now R satisfies (ACC) (resp. (DCC)) for ideals iff each Qi satisfies (ACC) (resp. (DCC)). But since each Qi is a vector space, (ACC) holds iff (DCC) holds (iff the dimension is finite). Theorem 215. (Akizuki-Hopkins) For a ring R, TFAE: (i) R is Artinian. (ii) R is Noetherian, and prime ideals are maximal. Proof. Assume R is Artinian. Then by Proposition X.X prime ideals are maximal, so it suffices to show that R is Noetherian. Let m1 , . . . , mn be the distinct maximal ∏n ∩n k ideals of R. For any k ∈ Z+ we have i=1 mk ⊂ ( i=1 mi ) . Applying Theorem i ∏n 213d), this shows that for sufficiently large k we have i=1 mk = 0. We can now i apply Lemma 214 to conclude that R is Artinian. Conversely, suppose R is Noetherian and zero-dimensional. Like any Noetherian ring, R has only finitely many minimal prime ideals, each of which is maximal by zero-dimensionality. Therefore ∩n N = i=1 mi is the nilradical of a Noetherian ring, hence a nilpotent ideal. As ∏n above, we deduce that for sufficiently large k we have i=1 mk = 0 and then we i apply Lemma 214 again to conclude that R is Artinian. Exercise 8.25: Let k be a field and A = k [{xi }∞ ] a polynomial ring over k in i=1 a countable infinite number of indeterminates. Let m = ({xi }) be the ideal of all polynomials with zero constant term, and put R = A/m2 . Show that R is a ring with a unique prime ideal which is not Noetherian (so also not Artinian). Exercise 8.26: Let n ∈ Z+ . Suppose R is a Noetherian domain with exactly n prime ideals. Must R be Artinian? Proposition 216. Let (R, m) be a Noetherian local ring. a) Either: (i) mk ̸= mk+1 for all k ∈ Z+ , or (ii) mk = 0 for some k . b) Moreover, condition (ii) holds iff R is Artinian. Proof. a) Suppose there exists k such that mk = mk+1 . By Nakayama’s Lemma, we have mk = 0. If p is any prime ideal of R, then mk ⊂ p, and taking radicals we have m ⊂ p, so p = m and R is a Noetherian ring with a unique prime ideal, hence an Artinian local ring. b) If R is Artinian, then (i) cannot hold, so (ii) must hold. Conversely, if (ii) holds then m is a nil ideal, hence contained in the intersection of all prime ideals of R, which implies that m is the only prime ideal of R, and R is Artinian by the Akizuki-Hopkins theorem. COMMUTATIVE ALGEBRA 139 Theorem 217. (Structure theorem for Artinian rings) Let R be an Artinian ring. ∏n a) There exist finitely many local Artinian rings Ri such that R ∼ i=1 Ri . = ∏m b) Moreover, the decomposition is unique in the sense that if R ∼ j =1 Sj is = another decomposition, then n = m and there exists a permutation σ of {1, . . . , n} such that Ri ∼ Sσ(i) for all i. = Proof. a) Let (mi )n be the distinct maximal ideals of R. We have seen that there i=1 ∏ n exists∩ ∈ Z+ such that i=1 mk = 0. By XX the ideals mk are pairwise comaximal, k i i ∏k so so i mk = i mi . Therefore by CRT the natural mapping i A→ n ∏A mk i i=1 is an isomorphism. Evidently this is a finite decomposition into a product of local Artinian rings, giving part a). b) The proof requires primary decomposition, so must be deferred to §10.5. Exercise: Let R be an Artinian ring. a) Show that every element of R is either a unit or a zero divisor. b) Show that R is its own total fraction ring. Exercise:36 For a ring R, TFAE: (i) R is semilocal [sic!], i.e., MaxSpec R is finite. (ii) R/ rad R is Artnian. 8.10. The Hilbert Basis Theorem. The following result shows in one fell swoop that the majority of the rings that one encounters in classical algebraic geometry and number theory are Noetherian. Theorem 218. (Hilbert Basis Theorem) Let R be a Noetherian ring. Then: a) The univariate polynomial ring R[t] is Noetherian. b) The formal power series ring R[[t]] is Noetherian. Proof. Seeking a contradiction, suppose J is an ideal of R[t] which is not finitely generated. We inductively construcct a sequence f0 , f1 , . . . , fn , . . . of elements of J and a sequence of ideals Jn = ⟨f0 , . . . , fn ⟩ of R[t] as follows: f0 = 0, and for all i ∈ N, fi+1 is an element of minimal degree in J \ Ji . Moreover, for all i ∈ Z+ let ai be the leading coefficient of fi , and let I be the ideal ⟨a1 , a2 , . . . , aN , . . .⟩ of R. However, R is Noetherian, so there exists N ∈ Z+ such that I = ⟨a1 , . . . , aN ⟩. In particular, there are u1 , . . . , uN ∈ R such that aN +1 = u1 a1 + . . . + uN aN . Dene g= N ∑ ui fi tdeg fN +1 −deg fi . i=1 Since g ∈ JN and fN +1 ∈ J \ JN , fN +1 − g ∈ J \ JN . Moreover, by construction g and fN +1 have the same degree and the same leading term, so deg fN +1 − g < deg fN +1 , hence fN +1 does not have minimal degree among polynomials in J \ JN , contradiction.37 36This exercise should be compared to Exercise 4.14, which gives a criterion for semilocality in terms of the quotient by the Jacobson radical. 37This argument is taken essentially verbatim from wikipedia. I like it a little better than the standard writeups! 140 PETE L. CLARK b) Put S = R[[t]], and let I be an ideal of S . For r ≥ 0, put let Ii be the ideal of R generated by the leading coefficients ai of all elements f = ai ti + ai+1 ti+1 + . . . with f ∈ I ∩ ti B . If f = ai ti + O(ti+1 ) ∈ Ii , then tf = ai ti+1 + O(ti+2 ) ∈ I ∩ ti+1 B , so Ii ⊂ Ii+1 for all i. By the assumed Noetherianity of R, there must therefore be an N such that IN = IN +k for all k ∈ N. For each 0 ≤ i ≤ N , let ai1 , . . . , aiM be a finite set of generators for Ii 38, and let gij ∈ S be a power series whose leading coefficient is aij . We claim that I = ⟨gij ⟩1≤i≤N,1≤j ≤M . Indeed, let f ∈ I . By choosing a suitable A-linear combination g0 of the g0j ’s we may arrange that f − g0 ∈ I ∩ tS , and similarly for 1 ≤ i ≤ N there exists gi ∈ Ii such that f− N ∑ gi ∈ I ∩ tN +1 S. i=0 Moreover, IN +1 = IN , so there exists gN +1 , an A-linear combination of the tgN j ’s, such that N +1 ∑ f− gi ∈ I ∩ tN +2 S. i=0 Continuing on in this way, we get an expression of f as an infinite power series, which can easily be regrouped to give an expression of the form f = g0 + . . . + gN + M ∑ hj g N j , j =1 QED. Exercise 8.27: Prove the converse of the Hilbert Basis Theorem: if R is a ring such that either R[t] or R[[t]] is Noetherian, then R is Noetherian. Corollary 219. A finitely generated algebra over a Noetherian ring is Noetherian. ∼ Proof. Let R be Noetherian and S a finitely generated R-algebra, so that S = R[t1 , . . . , tn ]/I for some n ∈ Z+ and some ideal I . By the Hilbert Basis Theorem (and induction), R[t1 , . . . , tn ] is Noetherian, hence so is its quotient ring S . Exercise 8.28: Show that for a ring R, TFAE: (i) R is Noetherian. (ii) For any n ≥ 1, R[[t1 , . . . , tn ]] is Noetherian. Exercise 8.29: Let k be a field, and consider the subring R = k [y, xy, x2 y, . . .] of k [x, y ]. Show that R is not Noetherian. Therefore, a subring of a Noetherian ring need not be Noetherian. Thinking that this ought to be the case is one of the classic “rookie mistakes” in commutative algebra. In general though, it is the exception rather than the rule that a nice property of a ring R is inherited by all subrings of R, and one gets used to this. 8.11. The Krull Intersection Theorem. 38The number of generators M should depend on i, but since there are only finitely many ideals in question, we may assume WLOG that each is generated by M generators for some M ∈ Z+ . COMMUTATIVE ALGEBRA 141 8.11.1. Preliminaries on Graded Rings. In the proof of the theorem of this section we will need a little fact about homogeneous polynomials. So here we discuss some bare rudiments of this theory by embedding it into its natural context: graded rings. The notion of graded ring is of the utmost importance in various applications of algebra, from algebraic geometry to algebraic topology and beyond. It would certainly be nice to give a comprehensive exposition of graded algebra but at the moment this is beyond the ambition of these notes, so we content ourselves with the bare minimum needed for our work in the next section. Let R be a ring, n ∈ Z+ , and denote by R[t] = R[t1 , . . . , tn ] the polynomial ring in n indeterminates over R. For a polynomial P = P (t) in several variables, we have the notion of the degree of P with respect to the variable ti : thinking of P as an element of R[t1 , . . . , ti−1 , ti+1 , . . . , tn ][ti ] it is just the largest m such that the coefficient of tm is nonzero, as usual. For any monomial term cI ti1 · · · tin we define n i 1 the total degree to be d = i1 + . . . + in . ∑ A nonzero polynomial P = I cI ti1 · · · tin is homogeneous if all of its monomial n 1 terms have the same total degree, and this common number is called the degree of the homogeneous polynomial P . By convention the zero polynomial is regarded as being homogeneous total degree d for all d ∈ N. A general polynomial P ∈ R[t] can be written as a sum of homogeneous poly∑∞ nomials P = d=0 Pd (t) with each Pd homogeneous of degree d (and of course Pd = 0 for all sufficiently large d). This sum is unique. One way to see this is to establish the following more structural fact: for any d ∈ N, let P [t]d be the set of all polynomials which are homogeneous of degree d. Then each P [t]d is an R-submodule of P [t] and we have a direct sum decomposition (11) P [t ] = ∞ ⊕ P [t ]d . d=0 Moreover, for all d1 , d2 ∈ N we have (12) P [t]d1 · P [t]d2 ⊂ P [t]d1 +d2 . In general, if R is a ring and S is an algebra admitting an R-module direct sum ⊕∞ decomposition S = d=0 Sd satisfying Sd1 · Sd2 ⊂ Sd1 +d2 , then we say that S is an (N)-graded R-algebra. Taking R = Z we get the notion of a graded ring. ⊕∞ Exercise: Let S = d=0 Sd be a graded R-algebra. Show that the R-submodule S0 is in fact an R-algebra. ⊕∞ Let S = d=0 Sd be a graded ring. We say that x ∈ S is homogeneous of degree d if x ∈ Sd . An ideal I of S is homogeneous if it has a generating set I = ⟨xi ⟩ with each xi a homogeneous element. Exercise: Let S be a graded R-algebra and let I be a homogeneous ideal of S . 142 PETE L. CLARK Show that S/I = ∞ ⊕ (Sd + I )/I d=0 and thus S/I is a graded R-algebra. Back to the case of polynomial rings. Lemma 220. Let S be a graded ring, let f1 , . . . , fn be homogeneous elements of S and put I = ⟨f1 , . . . , f⟩ , so that I is a homogeneous ideal of S . Let f ∈ I be any homogeneous element. Then there exist homogeneous elements g1 , . . . , g∈ R such that n ∑ f= gi fi i=1 and for all 1 ≤ i ≤ n deg gi = deg f − deg fi . Proof. Since f ∈ I , there exist X1 , . . . , Xn ∈ S such that f = X1 f1 + . . . + Xn fn . ∑ For each 1 ≤ i ≤ n, let Xi = j xi,j with deg xi,j = j be the canonical decomposition of Xi into a sum of homogeneous elements: i.e., deg xi,j = j . Then (13) f= ∞n ∑∑ xi,d−deg fi fi . d=0 i=1 Since f is homogeneous of degree deg(f ), only the d = deg(f ) in the right hand side of (13) is nonzero, so n ∑ f= xi,deg f −deg fi fi , i=1 qed. 8.11.2. The Krull Intersection Theorem. Theorem 221. Let R be a Noetherian ring, and I an ideal of R. Suppose there is ∩∞ an element x of R such that x ∈ n=1 I n . Then x ∈ xI . Proof. The following miraculously short and simple proof is due to H. Perdry [Pe04]. Suppose I = ⟨a1 , . . . , ar ⟩. For each n ≥ 1, since x ∈ I n there is a homogeneous degree n polynomial Pn (t1 , . . . , tr ) ∈ R[t1 , . . . , tr ] such that x = Pn (a1 , . . . , an ). By the Hilbert Basis Theorem (Theorem 218), R[t1 , . . . , tr ] is Noetherian. Therefore, definining Jn = ⟨P1 , . . . , Pn ⟩, there exists N such that JN = JN +1 . By Lemma 220 we may write PN +1 = QN P1 + . . . + Q1 PN , with Qi homogeneous of degree i > 0. Plugging in ti = ai for 1 ≤ i ≤ n, we get x = PN +1 (a1 , . . . , an ) = x (Q1 (a1 , . . . , an ) + . . . + QN (a1 , . . . , aN )) . Since each Qi is homogeneous of positive degree, we have Qi (a1 , . . . , an ) ∈ I , qed! COMMUTATIVE ALGEBRA 143 Corollary 222. Let R be a Noetherian ring and I an ideal of R. Suppose that either of the following holds: (i) R is a domain and I is a proper ideal. (ii) I ∩ contained in the Jacobson radical J (R) of R. is ∞ Then n=1 I n = 0. ∩∞ Proof. Either way, let x ∈ n=1 I n and apply Theorem 221 to obtain an element a ∈ I such that x = xa. Thus (a − 1)x = 0. Under assumption (i), we obtain either a = 1 – so I = R, contradicting the properness of I – or x = 0. Under assumption (ii), a ∈ J (R) implies a − 1 ∈ R× , so that we may multiply through by (a − 1)−1 , again getting x = 0. E a ∩xercise 8.30 (Su´rez-Alvarez): Exhibit an ideal I in a Noetherian ring such that ∞ I n {0}. (Hint: idempotents!) n=1 Exercise: Let R be the ring of all C ∞ functions f : R → R. Let m = {f ∈ R | f (0) = 0}. a) Show that m = xR is a maximal ideal of R. b) Show that for all n ∈ Z+ , mn = {f ∈ R | f (0) = f ′ (0) = . . . = f (n) (0)}. ∩∞ c) Deduce that n=1 mn is the ideal of all smooth functions with identically zero ∩∞ Taylor series expansion at x = 0. Conclude that n=1 mn ̸= 0. −1 d) Let f (x) = e x2 for x ̸= 0 and 0 for x = 0. Show that f ∈ f m. / e) Deduce that R is not Noetherian. ∪∞ 1 Exercise: Let R = n=1 C[[t n ]] be the Puiseux series ring. Show that R is a domain with a unique maximal ideal m and that for all n ∈ Z+ , mn = m. Deduce from KIT that R is not Noetherian. Remark: The preceding exercise will become much more routine when we study valuation rings in §17. In that language, one can show that if ∩ is a valuation ring R ∞ with divisible value group, then (R, m) is a local domain and n=1 mn = m. 8.12. Krull’s Principal Ideal Theorem. Theorem 223. (The Principal Ideal Theorem, a.k.a. Krull’s Hauptidealsatz) Let x be a nonunit in a Noetherian ring R, and let p be minimal among prime ideals containing x. Then p has height at most one. Remark: A prime p which is minimal among primes containing x will be called a minimal prime over x. Note that an equivalent condition is that p is a minimal prime in the quotient ring R/(x). Note also that if x is nilpotent, every prime of p contains x so the height of any minimal prime is 0. Our strategy of proof follows Kaplansky, who follows D. Rees. We need a preliminary result: Lemma 224. Let u and y be nonzero elements in a domain R. Then: a) The R-modules ⟨u, y ⟩/(u) and ⟨u2 , uy ⟩/(u2 ) are isomorphic. b) If we assume further that for all t ∈ R, tu2 ∈ (y ) implies tu ∈ (y ), then the R-modules (u)/(u2 ) and ⟨u2 , y ⟩/⟨u2 , uy ⟩ are isomorphic. 144 PETE L. CLARK Proof. a) The isomorphism is simply induced by multiplication by u. b) The module (u)/(u2 ) is cyclic with annihilator (u), and conversely any such module is isomorphic to R/(u). Moreover M := ⟨u2 , y ⟩/⟨u2 , uy ⟩ is also cyclic, being generated simply by y . Certainly u annihilates M , so it suffices to show that the annihilator is exactly (u). More concretely, given ky = au2 + buy , we must deduce that k ∈ (u). But we certainly have au2 ∈ (y ), so by hypothesis au ∈ (y ), say au = cy . Then ky = cuy + buy . Since 0 ̸= y in our domain R, we may cancel y to get k = (c + b)u ∈ (u). Proof of Krull’s Hauptidealsatz : Under the given hypotheses, assume for a contradiction that we have p2 p1 p. Note first that we can safely pass to the quotient R/p2 and thus assume that R is a domain. Dually, it does not hurt any to localize at p. Therefore we may assume that we have a Noetherian local domain R with maximal ideal m, an element x ∈ m, and a nonzero prime ideal, say p, with x ∈ p m, and our task is now to show that this setup is impossible. Now for the clever part: let 0 ̸= y be any element of p, and for k ∈ Z+ , let Ik denote the ideal of all elements t with txk ∈ (y ). Then {Ik }∞ is an ascending chain of ideals in the Noetherian ring R so must stabilize, k=1 say at k = n. In particular, tx2n ∈ (y ) implies txn ∈ (y ). Putting u = xn , we have tu2 ∈ (y ) implies (tu) ∈ (y ). Since m is a minimal prime over (x), the quotient ring T = R/(u2 ) has exactly one prime ideal, m, and is therefore, by the Akizuki-Hopkins Theorem, an Artinian ring, so that any finitely generated T -module has finite length. In particular, M := ⟨u, y ⟩/(u2 ), which can naturally be viewed as a T -module, has finite length, and hence so does its T -submodule M ′ := ⟨u2 , y ⟩/(u2 ). Put N = ⟨u2 , y ⟩⟨u2 , uy ⟩. Then ℓ(M ′ ) = ℓ(N ) + ℓ(⟨u2 , uy/(u2 )) = ℓ((u)/(u2 )) + ℓ(⟨u, y ⟩/(u)) = ℓ(M ); in the second equality we have used Lemma 224. The only way that M could have the same length as its submodule M ′ is if ⟨u, y ⟩ = ⟨u2 , y ⟩, i.e., if there exist c, d ∈ R such that u = cu2 + dy , or u(1 − cu) = −dy . Since u lies in the maximal ideal of the local ring R, 1 − cu ∈ R× , and thus u ∈ (y ) ⊂ p. But m is minimal over x and hence, being prime, also minimal over u = xn , contradiction! Corollary 225. With hypotheses as in Theorem 223, suppose that x is not a zerodivisor. Then any prime p which is minimal over x has height one. Exercise 8.31: Use the Akizuki-Hopkins theorem and Proposition 216 to give a proof of Corollary 225. As a second corollary of Theorem 223 we get the following striking structural result about primes in a Noetherian ring. First a piece of notation: for any elements x, y in a poset S we define the “interval” (x, y ) to be the set of all z ∈ S such that x < z < y . For prime ideals p and q, by (p, q) we understand that the ambient poset is the set of all prime ideals of R (which we are officially not yet allowed to call Spec R....), i.e., it is the set of all prime ideals strictly between p and q. Corollary 226. Let p ⊂ q be prime ideals in a Noetherian ring R. Then the interval (p, q) is either empty or infinite. COMMUTATIVE ALGEBRA 145 So if R is Noetherian with finitely many primes, its Krull dimension 0 or 1. Again we need a small preliminary result. Lemma 227. (Prime Avoidance) Let R be a ring, q an ideal of R and p1 , . . . , pn ∪ prime ideals. If q ⊂ i pi , then q ⊂ pi for some i. Proof. We go by induction on n, the case n = 1 being trivial. Assume the result for n − 1. Then if q were contained in the union of any n − 1 of the primes pi , then we would be done by induction, so we may assume the existence of, for all 1 ≤ i ≤ n, ∪ an element xi ∈ pi ∩ (q \ j ̸=i pi ). If n = 2, put y = x1 + x2 ; then y is an element of q but not of either p1 or p2 , a contradiction. In general, put y = x1 + x2 x3 · · · xn ; then if y ∈ p1 then so is x2 · · · xn and since p1 is prime, for some i > 1, xi ∈ p1 , contradiction. Similarly if y ∈ pi for i > 1 then x1 ∈ pi , contradiction. Exercise 8.32 (Kaplansky): Generalize the lemma to the case in which all but at most two of the ideals pi are prime. Proof of Corollary 226: As usual, by correspondence we may pass to R/p and therefore assume WLOG that p = 0. Suppose that for some n ≥ 1, [0, q] = {p1 , .∪. , pn }. . ∪n n According to Lemma 227 we cannot then have q ⊂ i=1 pi , so choose x ∈ q \ i=1 pi . Then q is a prime of R, of height at least 2, which is minimal over (x), contradicting Theorem 223. Theorem 228. (Generalized Principal Ideal Theorem) Let R be a Noetherian ring, and let I = ⟨a1 , . . . , an ⟩ be a proper ideal of R. Suppose that I can be generated by n elements. Let p be a minimal element of the set of all prime ideals containing I . Then p has height at most n. Proof. As usual, we may localize at p and suppose that R is local with p as its maximal ideal. Suppose to the contrary that there exists a chain p = p0 p1 . . . pn+1 . Because R is Noetherian, we may arrange for (p1 , p) = ∅. Because p is minimal over I , I cannot be contained in p1 ; without loss of generality we may suppose that a1 is not in p1 . Put J := ⟨p1 , a1 ⟩; then J strictly contains p1 so p is the unique prime of R containing J . So the ring R/J is an Artin local ring, and then by Proposition 216 for sufficiently large k we have pk ⊂ J . Then by taking t to be sufficiently large we can write, for 2 ≤ i ≤ n, at = ci a1 + bi , ci ∈ R, bi ∈ p1 . i Put K = ⟨b2 , . . . , bn ⟩ ⊂ p1 . Since the height of p1 exceeds n − 1, by induction on n we may assume that p1 properly contains a prime ideal Q which contains J . Now the ideal Q′ := ⟨a1 , Q⟩ contains some power of each ai and therefore p is the unique prime ideal containing Q′ . Therefore in the quotient domain R/Q, the prime p/Q is minimal over the principal ideal Q′ /Q, so by Krull’s Hauptidealsatz p/Q has height 1. But on the other hand we have p/Q p1 /Q 0, a contradiction. 8.13. The Dimension Theorem. The following is a very basic theorem about Noetherian rings: Theorem 229. (Dimension Theorem) Let R be a Noetherian ring, and R[t] be the univariate polynomial ring over R. Then dim R[t] = dim R + 1. 146 PETE L. CLARK Remark: For a non-Noetherian ring R, one has the inequalities dim R + 1 ≤ dim R[t] ≤ 2 dim R[t] + 1, and indeed there are rings for which dim R[t] = 2 dim R + 1. Of course, an immediate induction argument gives: Corollary 230. Let k be a field. Then dim k [t1 , . . . , tn ] = n. At this time we do not aspire to give a proof of Theorem 229. Neither will we use it formally in the sequel, i.e., the proof of no future result will depend on it. Nevertheless it is such a simple and natural fact that to deprive the reader of it would be a serious blow against developing any intuition for the notion of dimension of a Noetherian ring. 8.14. The Artin-Tate Lemma. Theorem 231. (Artin-Tate [AT51]) Let R ⊂ T ⊂ S be a tower of rings such that: (i) R is Noetherian, (ii) S is finitely generated as an R-algebra, and (iii) S is finitely generated as a T -module. Then T is finitely generated as an R-algebra. Proof. Let x1 , . . . , xn be a set of generators for S as an R-algebra, and let ω1 , . . . , ωm be a set of generators for S as a T -module. For all 1 ≤ i ≤ n, we may write ∑ (14) xi = aij ωj , aij ∈ T. j Similarly, for all 1 ≤ i, j ≤ m, we may write ∑ bijk ωk , bijk ∈ T. (15) ωi ωj = i,j,k Let T0 be the R-subalgebra of T generated by the aij and bijk . Since T0 is a finitely generated algebra over the Noetherian ring R, it is itself a Noetherian ring by the Hilbert Basis Theoerem. Now each element of S may be expressed as a polynomial in the xi ’s with Rcoefficients. Making substitutions using (14) and then (15), we see S is generated as a T0 -module by ω1 , . . . , ωm , and in particular that S is a finitely generated T0 module. Since T0 is Noetherian, the submodule T is also finitely generated as a T0 -module. This immediately implies that T is finitely generated as a T0 -algebra and then in turn that T is finitely generated as an R-algebra, qed! 9. Boolean rings 9.1. Definition and first properties of Boolean rings. Let R be a ring, not necessarily commutative, but with a multiplicative identity 1, with the property that x2 = x for all x ∈ R. Then (1 + 1) = (1 + 1)2 = 1 + 1 + 1 + 1, so 1 + 1 = 0. It follows that −x = x for all x ∈ R. Moreover for any x, y ∈ R, (x + y ) = (x + y )2 = x2 + xy + yx + y 2 = x + y + xy + yx, so xy + yx = 0 or xy = yx. Therefore such a ring is necessarily commutative. COMMUTATIVE ALGEBRA 147 With this remark in mind, define a Boolean ring to be a commutative ring with identity such that x2 = x for all elements x. Exercise 9.1: Show that the group of units of a Boolean ring is trivial. Exercise 9.2: a) Show that any quotient ring of a Boolean ring is Boolean. b) Show that any subring of a Boolean ring is Boolean. Exercise 9.3: Show that a Boolean ring is absolutely flat. 9.2. Boolean Algebras. A Boolean ring is an object of commutative algebra. It turns out that there is a completely equivalent class of structures of an order-theoretic nature, called Boolean algebras. In some ways the concept of a Boolean algebra is more intuitive and transparent – e.g., starting directly from the definition, it is perhaps easier to give examples of Boolean algebras. Moreover it is not at all difficult to see how to pass from a Boolean ring to a Boolean algebra and conversely. A Boolean algebra is a certain very nice partially ordered set (B, ≤). Recall that for any partially orderet set B and any subset S , we have the notion of the supremum sup S and the infimum inf S . To define these it is convenient to extend the inequality notation as follows: if S, T are subsets of B , we write S<T to mean that for all s ∈ S and t ∈ T , s < t, and similarly S≤T to mean that for all s ∈ S and t ∈ T , s ≤ t. Then we say that z = sup S if S ≤ z and if w is any element of B with S ≤ w, then z ≤ w. Similarly z = inf S if z ≤ S and if w is any element of B with w ≤ S then w ≤ Z . For a given subset S , neither sup S nor inf S need exist, but if either exists it it is plainly unique. In particular if sup ∅ exists, it is necessarily a bottom element, called 0, and if inf ∅ exists, it is necessarily a top element called 0. A partially ordered set (L, ≤) is called a lattice if for all x, y ∈ L, sup{x, y } and inf {x, y } both exist. We give new notation for this: we write x ∨ y := sup{x, y }, the join of x and y and x ∧ y := inf {x, y }, the meet of x and y . A lattice is said to be bounded if it contains a bottom element 0 and a top element 1: equivalently, sup S and inf S exist for every finite subset S . Exercise 9.4: Let L be a lattice containing 0 and 1, and let x ∈ L. Then: a) x ∨ 1 = 1, b) x ∧ 1 = x, c) x ∨ 0 = x, d) x ∧ 0 = 0. 148 PETE L. CLARK A lattice L is complemented if it has a bottom element 0, a top element 1, and for each x ∈ L there exists y ∈ L such that x ∨ y = 1, x ∧ y = 0. A lattice is distributive if ∀x, y, z ∈ L, (x ∨ y ) ∧ z = (x ∧ z ) ∨ (y ∧ z ), (x ∧ y ) ∨ z = (x ∨ z ) ∧ (y ∨ z ). Proposition 232. Let L be a distributive complemented lattice. Then for all x ∈ L, the complement of x is unique. Proof. Suppose that y1 and y2 are both complements to x, so x ∨ y1 = x ∨ y2 = 1, x ∧ y1 = x ∧ y2 = 0. Then y2 = 1 ∧ y2 = (x ∨ y1 ) ∧ y2 = (x ∧ y2 ) ∨ (y1 ∧ y2 ) = 0 ∨ (y1 ∧ y2 ) = y1 ∧ y2 , so y2 ≤ y1 . Reasoning similarly, we get y1 ≤ y2 , so y1 = y2 . By virtue of Proposition 232 we denote the complement of an element x in a distributive complemented lattice as x∗ . Exercise 9.5: Show that for every element x of a distributive complemented lattice we have (x∗ )∗ = x. A Boolean algebra is a complemented distributive lattice with 0 ̸= 1. Exercise 9.6: Show that DeMorgan’s Laws hold in any Boolean algebra B : for all x, y ∈ B , we have a) (x ∧ y )∗ = x∗ ∨ y ∗ and b) (x ∨ y )∗ = x∗ ∧ y ∗ . The shining example of a Boolean algebra is the powerset algebra 2S for a nonempty set S . In the special case in which |S | = 1, we denote the corresponding Boolean algebra (the unique totally ordered set on two elements) simply as 2. Not every Boolean algebra is isomorphic to a power set Boolean algebra. Example: Let S be a set, and let Z (S ) ⊂ 2S be the collection of all finite and cofinite subsets of S . It is easily checked that (Z (S ), ⊂) ⊂ (2S , ⊂) is a sub-Boolean algebra. However, |Z (S )| = |S |, so if |S | = ℵ0 , then Z (S ) is not isomorphic to any power set Boolean algebra. Boolean algebras form a full subcategory of the category of partially ordered sets. In other words, we define a morphism f : B → B ′ of Boolean algebras simply to be an isotone (or order-preserving) map: ∀x, y ∈ B, x ≤ y =⇒ f (x) ≤ f (y ). One can (and sometimes does, e.g. for model-theoretic purposes) also axiomatize Boolean algebras as a structure (B, ∨, ∧, ∗, 0, 1), the point being that x ≤ y iff x ∨ y = y iff x ∧ y = x, so the partial ordering can be recovered from the wedge or the join. COMMUTATIVE ALGEBRA 149 Proposition 233. The category of Boolean rings is equivalent to the category of Boolean algebras. In other words, we can define a functor F from Boolean rings to Boolean algebras and a functor G from Boolean algebras to Boolean rings such that for every Boolean ring R, R is naturally isomorphic to G(F (R)) and for every Boolean algebra B , B is naturally isomorphic to F (G(B )). Let us sketch the basic construction, leaving the details to the reader. Suppose first that R is a Boolean ring. Then we associate a Boolean algebra F (R) with the same underlying set as R, endowed with the following operations: ∀x, y ∈ R, x ∧ y = xy, x∗ = 1 − x. We should also of course define the join operation, but the point is that it is forced on us be DeMorgan’s Laws: x ∨ y = (x∗ ∧ y ∗ )∗ = x + y − xy. Exercise 9.7: Check that (F (R), ∧, ∨, ∗) is indeed a Boolean algebra, and that the bottom element 0 in F (R) (resp. the top element 1) is indeed the additive identity 0 (resp. the multiplicative identity 1). Conversely, suppose that we have a Boolean algebra (B, ∧, ∨, ∗). Then we define a Boolean ring G(B ) on the same underlying set B , by taking x + y := (x ∧ y ∗ ) ∨ (y ∧ x∗ ) xy := x ∧ y. Note that the addition operation corresponds to the Boolean operation “exclusive or” or, in more set-theoretic language, symmetric difference x∆y . Exercise 9.8: check that (G(B ), +, ·) is indeed a Boolean ring with additive identity the bottom element 0 of B and multiplicative identity the top element 1 of B . Exercise 9.9: a) Let R be a Boolean ring. Show that the identity map 1R on R is an isomorphism of Boolean rings R → G(F (R)). b) Let B be a Boolean algebra. Show that the identity map 1B on B is an isomorphism of Boolean algebras B → G(F (B )). Exercise 9.10: Let X be a nonempty set, let BX be the Boolean algebra of subsets of X , partially ordered by inclusion. Show that the corresponding Boolean ring may be identified with the ring 2X of all functions from X to F2 under pointwise addition and multiplication. Exercise 9.11: Show that every finite Boolean algebra is isomorphic to a powerset algebra. Conclude that every finite Boolean ring R is isomorphic to the ring of binary functions on a finite set of cardinality log2 (#R). In fact, try to show this both on the Boolean algebra side and on the Boolean ring side. (Hint for the Boolean ring side: use the decomposition into a direct product afforded by an 150 PETE L. CLARK idempotent element.) Exercise 9.12: Show that an arbitrary direct product of Boolean algebras (or, equivalently, Boolean rings) is a Boolean algebra (or...). As we saw above, cardinality considerations already show that not every Boolean algebra is the power set Boolean algebra, and hence not every Boolean ring is the full ring of binary functions on some set X . However, in view of results like Cayley’s theorem in basic group theory, it is a reasonable guess that every Boolean algebra is an algebra of sets, i.e., is a sub-Boolean algebra of a power set algebra. We proceed to prove this important result on the Boolean ring side. 9.3. Ideal Theory in Boolean Rings. Proposition 234. Let R be a Boolean ring. a) For all x ∈ N and all n ≥ 2, xn = x. b) A Boolean ring is reduced, i.e., has no nonzeronilpotent elements. c) Every ideal in a Boolean ring is a radical ideal. Proof. a) The case n = 2 is the definition of a Boolean ring, so we may assume n ≥ 3. Assume the result holds for all x ∈ R and all 2 ≤ k < n. Then xn = xn−1 x = x · x = x. b) If x ∈ R is such that xn = 0 for some positive integer n, then either n = 1 or n ≥ 2 and xn = x; either way x = 0. c) Let I be an ideal in the Boolean ring R. Then I = rad(I ) iff R/I is reduced, but R/I is again a Boolean ring and part b) applies. The ring Z/2Z is of course a Boolean ring. It is also a field, hence certainly a local ring and an integral domain. We will shortly see that it is unique among Boolean rings in possessing either of the latter two properties. Proposition 235. a) The only Boolean domain is Z/2Z. b) Every prime ideal in a Boolean ring is maximal. Proof. a) Let R be a Boolean domain, and let x be an element of R. Then x(x − 1) = 0, so in a domain R this implies x = 0 or x = 1, so that R ∼ Z/2Z. = b) If p is a prime ideal in the Boolean ring R, then R/p is a Boolean domain, hence – by part a) – is simply Z/2Z. But this ring is a field, so p is maximal. Proposition 236. Let R be a local Boolean ring. Then R ∼ Z/2Z. = Proof. Let m be the unique maximal ideal of R. By Proposition 235, m is moreover the unique prime ideal of R. It follows from Proposition 121d) that m = nil R is the set of all nilpotent elements, so by Proposition 234b) m = 0. Thus R is a field and thus, by Proposition 235 must be Z/2Z. Exercise 9.13: Let R be a Boolean ring, let I be an ideal of R. a) Show that x, y ∈ I =⇒ x ∨ y ∈ I . b) Show in fact that ⟨x, y ⟩ = ⟨x ∨ y ⟩. c) Deduce that any finitely generated ideal of R is principal. Thus, for a Boolean ring R, all ideals of R are principal iff R is Noetherian. But in fact very few Boolean rings are Noetherian: we have already seen them all in Exercise X.X. COMMUTATIVE ALGEBRA 151 Proposition 237. For a Boolean ring R, the following conditions are equivalent: (i) R is finite. (ii) R is Noetherian. (iii) R has finitely many maximal ideals. If these equivalent conditions hold, then R ∼ (Z/2Z)n , with n = log2 #R. = Proof. That (i) =⇒ (ii) is clear. (ii) =⇒ (iii): Since R is Noetherian and prime ideals are maximal, by the AkizukiHopkins Theorem (Theorem 215) R is Artinian. Thus by Theorem 217 ⊕ is a finite R n product of local Boolean rings, and thus finally by Proposition 236 R ∼ i=1 Z/2Z. = (iii) =⇒ (i): Suppose that R has precisely N < ∞ maximal ideals and suppose for a contradiction that R is infinite. Then we may choose x ∈ R \{0, 1}, i.e., a nontrivial idempotent. This leads to a direct product decomposition R = xR × (1 − x)R. Here xR and (1 − x)R are subrings of R, hence at least one of them is an infinite Boolean ring. It follows that this decomposition process can be continued indefinitely, or ⊕N +1 more precisely until we get to the point of writing R = i=1 Ri as a product of N + 1 Boolean rings. For i = 1, . . . , N + 1, let mi be a maximal ideal of Ri . Then ∏ for i = 1, . . . , N + 1, Mi = j ̸=i Rj × mi are distinct maximal ideals of R, which thus has at least N + 1 maximal ideals, contradiction. Lemma 238. Let m be an ideal in the Boolean ring R. TFAE: (i) m is maximal. (ii) For all x ∈ R, either x ∈ R or 1 − x ∈ R (and not both!). Proof. (i) =⇒ (ii): Of course no proper ideal in any ring can contain both x and 1 − x for then it would contain 1. To see that at least one must lie in m, it is certainly no loss to assume that x is neither 0 nor 1, hence R = xR × (1 − x)R. Recall that in a product R1 × R2 of rings, every ideal I is itself a product I1 × I2 , where Ii is an ideal of Ri . Then R/I ∼ R1 /I1 × R2 /I2 . So I is prime iff R/I is a = domain iff either (i) I1 is prime in R1 and I2 = R2 or (ii) I1 = R1 and I2 is prime in R2 . In particular, every maximal ideal of R1 × R2 contains either R1 or R2 : done. (i) =⇒ (ii):39 We prove the contrapositive: if m is not maximal, there exists a maximal ideal M properly containing m. Let x ∈ M \ m. Since M is proper, it does not also contain 1 − x, hence neither does the smaller ideal m. Exercise 9.14 (Kernel of a homomorphism): Let f : R → F2 be a homomorphism of Boolean rings. a) Show that Ker f is ∨-closed: if x, y ∈ Ker f , then x ∨ y ∈ Ker f . b) Show that Ker f is downward-closed: if x ∈ Ker f and y ≤ x, then y ∈ Ker f . c) Explain why parts a) and b) are equivalent to showing that Ker f is an ideal of the Boolean ring R. d) Show that Ker f is in fact a maximal ideal of R. e) Conversely, for every maximal ideal m of R, show that R/m = F2 and thus the quotient map q : R → R/m is a homomorphism from R to F2 . Exercise 9.15 (Shell of a homomorphism): Let f : R → F2 be a homomorphism of Boolean rings. Define the shell Sh f to be f −1 (1). a) Show that Sh f is wedge-closed: if x, y ∈ Sh f , so is x ∧ y . b) Show that Sh f is upward-closed: if x ∈ Sh f and x ≤ y , then y ∈ Sh f . 39Note that this direction holds in any ring. 152 PETE L. CLARK c) A nonempty, proper subset of a Boolean algebra which is wedge-closed and upward-closed is called a filter, so by parts a) and b) Sh f is a filter on B . Show that in fact it is an ultrafilter on B , i.e., that it is not properly contained in any other filter. (Suggestion: use Lemma 238.) d) Show that every ultrafilter on B is the shell of a unique homomorphism of Boolean algebras f : B → F2 . 9.4. The Stone Representation Theorem. Let R be a Boolean ring. We would like to find an embedding of R into a Boolean ring of the form 2X . The key point of course, is to conjure up a suitable set X . Can we find any clues in our prior work on Boolean rings? indent Well, finite Boolean rings we understand: the proof of Proposition 237 gives ⊕ us that every finite Boolean ring is of the form i∈X Z/2Z, where the elements ⊕ of X correspond to the maximal ideals of R. The isomorphism i∈X Z/2Z ∼ 2X = amounts to taking each element x of R and recording which of the maximal ideals it lies in: namely, x lies in the ith maximal ideal mi of all elements having a zero in the ith coordinate iff its image in the quotient R/mi = Z/2Z is equal to 0. This motivates the following construction. For any Boolean ring, let M (R) be the set of all maximal ideals of R, and define a map E : R → M (R) by letting E (x) be the set of maximal ideals of R which do not contain x. This turns out to be very fruitful: Theorem 239. (Stone Representation Theorem) Let R be a Boolean ring and M (R) the set of maximal ideals. The map E : R → 2M (R) which sends an element x of R to the collection of all maximal ideals of R which do not contain x is an injective homomorphism of Boolean rings. Therefore R is isomorphic to the Boolean ring associated to the algebra of sets E (R) ⊂ 2M (R) . In particular this shows that every Boolean algebra is an algebra of sets. Proof. Step 1: We check that the map E is a homomorphism of Boolean algebras. Above we saw E (0) = ∅; also E (1) = M (R). Also, for x, y ∈ R, E (xy ) is the set of maximal ideals which do not contain xy ; since maximal ideals are prime this is the set of maximal ideals which contain neither x nor y , i.e., E (x) ∩ E (y ) = E (x) · E (y ). Finally, E (x) + E (y ) = E (x) ∆ E (y ) is the set of maximal ideals which contain exactly one of x and y , whereas E (x + y ) is the set of maximal ideals not containing x + y . For m ∈ M (R), consider the following cases: (i) x, y ∈ m. Then m is not in E (x) ∆ E (y ). On the other hand x + y ∈ m, so m is not in E (x + y ). (ii) Neither x nor y is in m. Certainly then m is not in E (x) ∆ E (y ). On the other hand, remembering that R/m ∼ Z/2Z, both x and y map to 1 in the quotient, so = x + y maps to 1 + 1 = 0, i.e., x + y ∈ m, so m is not in E (x + y ). (iii) Exactly one of x and y lies in m. Then m ∈ E (x) ∆E (y ) and as above, x + y maps to 1 in R/m, so x + y is not in m and m ∈ E (x + y ). Step 2: We show that the map E is injective. In other words, suppose we have two elements x and y of R such that a maximal ideal m of R contains x iff it contains COMMUTATIVE ALGEBRA y . Then (x) = rad(x) = ∩ m∈M (R) | x∈ m m= ∩ 153 m = rad(y ) = (y ). m∈M (R) | y ∈ m So there exist a, b ∈ R with y = ax, x = by , and then x = by = by 2 = xy = ax2 = ax = y. Let us comment a bit on the proof. Although Step 1 is longer, it is clearly rather routine. They key is of course that E gives an embedding, which as we saw is equivalent to the much gutsier statement that an element of a Boolean ring is entirely determined by the family of maximal ideals containing it. From the standpoint of the more “conventional” rings one encounters in number theory and algebraic geometry, this is a very strange phenomenon. First of all, it can only be true in a ring R which has trivial unit group, since if u is a nontrivial unit of course we will not be able to distinguish 1 and u using ideals! Moreover it implies that every prinicpal ideal is radical, which is impossible in any integral domain. Finally it implies that every radical ideal is the intersection of the maximal ideals which contain it, a property which we will meet later on the course: such rings are called Jacobson rings. 9.5. Boolean spaces. For reasons which I hope will shortly be made clear, we will now digress a bit to talk (not for the first or last time!) about topological spaces. Following Bourbaki, for us compact means quasi-compact and Hausdorff. Further a locally compact space is a Hausdorff space in which each point admits a local base of compact neighborhoods. A subset of a topological space is clopen if it is both closed and open. Recall that a topological space X is totally disconnected if the maximal connected subsets of X are the singleton sets {x}. Note that a totally disconnected space is necessarily separated (older terminology that I am not fond of: T1 ): i.e., singleton sets are closed. Indeed, the closure of every connected set is connected, so the closure of a non-closed point would give a connected set which is larger than a point. On the other hand a space X is zero-dimensional if it admits a base of clopen sets. Proposition 240. Let X be a locally compact space. Then X is totally disconnected iff it is zero-dimensional. Proof. Exercise! (This is not used in the sequel.) A space X is called Boolean40 if it is compact and zero-dimensional; in particular a Boolean space admits a base for the topology consisting of compact open sets. Exercise 9.16: a) A finite space is Boolean iff it is discrete. b) A Boolean space is discrete iff it is finite. c) An arbitrary direct product of Boolean spaces is Boolean. d) The usual Cantor space is homeomorphic to a countably infinite direct product 40There are many synonyms: e.g. Stone space, profinite space. 154 PETE L. CLARK of copes of a discrete, two-point space and thus is a Boolean space. Exercise 9.17: Show that a topological space is Boolean iff it is homeomorphic to an inverse limit of finite, discrete spaces. To every topological space X we may associate a Boolean algebra: namely, the subalgebra of 2X consisting of compact open subsets. Thus in particular we may associate a Boolean ring, say C (X ), the characteristic ring of X . (We also ourselves to pass between C (X ) and the associated Boolean algebra on the same set and call the latter the characteristic algebra.) Exercise 9.18: Show that the assignment X → C extends to a contravariant functor from the category of topological spaces to the category of Boolean rings. (In other words, show that a continuous map f : X → Y of topological spaces induces a “pullback” homomorphism C (f ) : C (Y ) → C (x) of Boolean rings.) If X is itself a Boolean space, then the characteristic algebra C (X ) is indeed characteristic of X in the following sense. Proposition 241. Let X be a Boolean space, and let A be a Boolean algebra of subsets of X which is also a base for the topology of X . Then A = C (X ). Proof. By hypothesis the elements of A are open sets in X . Moreover, since A is closed under complementation, the elements are also closed. Thus A ⊂ C (X ). Conversely, suppose Y ∈ C (X ). Since Y is open and A is a base for the topology on X , for each y ∈ Y there is Ay ∈ A with y ∈ Ay ⊂ Y . Thus {Ay }y∈Y is an open cover for Y . But Y is also closed in a ∪ compact space hence itself compact, so n we may extract a finite subcover, say Y = i=1 Ayi . Since A is a subalgebra, it is closed under finite unions, so Y ∈ A. Thus C (X ) ⊂ A. To every Boolean ring R we may associate a Boolean space: namely there is a natural topology on M (R), the set of maximal ideals of R, with respect to which M (R) is a Boolean space. This topology can be described in many ways. First Approach: by the Stone Representation Theorem we have an embedding R → 2M (R) and thus every element x ∈ R determines a function x : M (R) → F2 . We may endow F2 with the discrete topology (what else?) and then give M (R) the initial topology for the family of maps {x : M (R) → F2 }x∈R , that is the finest topology which makes each of these maps continuous. Here is a more concrete description of this initial topology: for each x ∈ R, put Ux = {m ∈ M (R) | x ∈ m} / Vx = {m ∈ M (R) | x ∈ m}. Then the topology in question is the one generated by {Ux , Vx }x∈X . Second Approach: for any Boolean ring R, to give a maximal ideal m of R is equivalent to giving a homomorphism of Boolean rings f : R → F2 . Namely, to a maximal ideal m we associate the quotient map, and to f : R → F2 we associate the kernel f −1 (0). In this way we get an embedding ι : M (R) → 2R . Now we endow COMMUTATIVE ALGEBRA 155 each copy of F2 with the discrete topology and 2R with the product topology: this makes it into a Boolean space. Lemma 242. The image ι(M (R)) of ι is a closed subspace of 2R . Exercise 9.19: Prove Lemma 242. Thus if we endow M (R) with the topology it inherits via the embedding ι, it is itself a Boolean space. Exercise 9.20: Show that the topology on M (R) defined via Lemma 242 coincides with the initial topology on M (R) defined above. It turns out to be important to consider a distinguished base for the topology on M (R), which we now define. Since for a prime ideal m we have xy ∈ m ⇐⇒ / x ∈ m andy ∈ m, we have for all x, y ∈ R that / / Ux ∩ Uy = Uxy . Moreover, by Lemma 238, M (R) \ Vx = Ux . It follows that the {Ux }x∈X form a base of clopen sets for the topology on M (R). To show the utility of this base, let us use it to show directly that M (R) is a Boolean space. Hausdorff: Let m1 and m2 be distinct maximal ideals of R. Choose x ∈ m2 \ m1 , so by Lemma 238 1 − x ∈ m1 \ m2 . Thus m1 ∈ Ux m2 ∈ U1−x and Ux ∩ U1−x = Ux(1−x) = U0 = ∅, so we have separated m1 and m2 by open sets. Quasi-compact: As is well-known, it is enough to check quasi-compactness of a space using covers by elements of any fixed base. We certainly have a preferred base here, namely {Ux }x∈X , so let’s use it: suppose that we have a collection ∪ {xi }i∈I such that i∈I Uxi = M (R). Now again (and not for the last...) we exploit the power of DeMorgan: ∪ ∪ ∩ M (R ) = Uxi = (M (R) \ Vxi ) = M (R) \ V xi , ∩ i i i so that i Vxi = ∅. This means that there is no maximal ideal containing every xi . But that means that the ideal generated by the xi ’s contains 1:∩there exists a ∑ finite subset J ⊂ I and aj ∈ R such that j aj xj = 1, and thus j ∈J Vxj = ∅: ∪ equivalently j ∈J Uxj = M (R). Thus we have shown that the correspondence R → M (R) associates to every Boolean ring a Boolean topological space, its Stone space. Exercise 9.21: Show that the assignment R → M (R) extends to a functor from the category of Boolean rings to the category of Boolean spaces. 156 PETE L. CLARK 9.6. Stone Duality. Theorem 243. (Stone Duality): The functors C and M give a duality between the category of Boolean spaces and the category of Boolean algebras. More concretely: a) For every Boolean algebra B , the map B → C (M (B )) given by x ∈ B → Ux is an isomorphism of Boolean algebras. b) For every Boolean space X , the map m : X → M (C (X )) given by x ∈ X → mx := {U ∈ C (X ) | x ∈ U } is a homeomorphism of Boolean spaces. / Proof. a) The map e : x ∈ B → Ux ∈ 2M (B ) is nothing else than the embedding e of the Stone Representation Theorem. In particular it is an embedding of Boolean algebras. Its image e(B ) is a subalgebra of the characteristic algebra of the Boolean space M (B ) which is, by definition, a base for the topology of M (B ). By Proposition 241 we have e(B ) = C (M (B )) so e is an isomorphism of Boolean algebras. b) First we need to show that mx is a maximal ideal in the characteristic ring C (X ). It seems more natural to show this on the Boolean algebra side, i.e., to show that mx is downward closed and union-closed. Indeed, U ∈ mx means x ∈ U , so if V ⊂ X / then certainly x ∈ V , i.e., V ∈ mx ; moreover, U, V ∈ mx ⇐⇒ x ∈ U and x ∈ / / / V ⇐⇒ x ∈ U ∪ V ⇐⇒ U ∪ V ∈ mx . Thus mx is an ideal of C (X ). Applying / Lemma 238, one easily sees that is maximal, so the map m is well-defined. indent The injectivity of m follows immediately from the Hausdorff property of X . Surjectivity: Let m ∈ M (C (X )). By Exercise X.X, we may identify m with a − homomorphism of Boolean algebras fm : C (X ) → F2 . Let F = fm 1 (1) be the shell of fm , an ultrafilter on the Boolean algebra of sets C (X ). In particular F is wedge-closed, i.e., it is a family of clopen subsets of the compact space X satisfy∩ ing the finite intersection property. Therefore there exists x ∈ U ∈F U . On the other hand, the collection Fx of all clopen sets in X containing x is also a filter on C (X ) with F ⊂ Fx . But since F is an ultra filter – i.e., a maximal filter – we have F = Fx . Thus m and Fx are respectively the kernel and shell of the homomorphism f : C (X ) → F2 , so m = C (X ) \ Fx = {U ∈ C (X ) | x ∈ U } = m(x). / Finally, since m is surjective, we have that for each A ∈ C (X ), {U ∈ M (C (X )) | A ∈ U } = {m(x) | x ∈ A}, so that m maps the base C (X ) for the topology on X onto the base C (M (C (X ))). Exercise 9.22: Let X be a topological space, and let C (X, 2) be the ring of all continuous functions f : X → F2 (F2 being given the discrete topology). a) Show that C (X, 2) is a Boolean ring. b) Suppose that X = M (R) is the maximal ideal space of the Boolean ring R. Show that C (X, 2) is canonically isomorphic to R itself. Thus every Boolean ring is the ring of continuous Boolean-valued functions on its Stone space of maximal ideals. 9.7. Topology of Boolean Rings. Proposition 244. Let R be a Boolean ring and m a maximal ideal of R. TFAE: (i) m is an isolated point in the Stone space M (R). (ii) m = Rx is a principal ideal. COMMUTATIVE ALGEBRA 157 Exercise 9.23: Prove Proposition 244. A Boolean ring R is atomic if for every x ̸= 1 there exists a principal maximal ideal m with x ∈ m. Exercise 9.24: For any nonempty set S , show that 2S = ∏ s ∈S Z/2Z is atomic. A Boolean ring is called atomless if it contains no maximal principal ideals. Exercise 9.25: Show that a Boolean algebra B is atomless if for all x ∈ B , if x < 1, there exists y ∈ B with x < y < 1. Proposition 245. A Boolean ring R is atomless iff its Stone space M (R) is perfect, i.e., without isolated points., Exercise 9.26: Prove Proposition 245. Corollary 246. Any two countably infinite atomless Boolean rings are isomorphic. Exercise 9.27: Prove Corollary 246. (Suggestion: show that the Stone space of any countably infinite atomless Boolean ring is isomorphic to the Cantor set.) Exercise 9.28 (for those who know some model theory): a) Show that there is a first order theory in the language (∨, ∧, ∗, 0, 1) whose models are precisely the atomless Boolean algebras. b) Use Vaught’s Test to show that this theory is complete. Exercise 9.29: Let S be a nonempty set and consider the Boolean ring R = 2S = ∏ s∈S Z/2Z. a) Show that there is a natural bijective correspondence between elements of S and principal maximal ideals of R. b) Deduce that there is an embedding ι : S → M (R) such that the induced topology on S is discrete. c) Show that ι(S ) is dense in M (R). (Hint: R is atomic.) d) Show that ι is a homeomorphism iff S is finite. e)* Show that ι is none other than the Stone-Cech compactification of the discrete space X . 10. Primary Decomposition 10.1. Some preliminaries on primary ideals. Recall that a proper ideal q of a ring R is primary if for all x, y ∈ R, xy ∈ q implies x ∈ q or y n ∈ q for some n ∈ Z+ . Exercise 10.1: a) Show that a prime ideal is primary. (Trivial but important!) b) Show that an ideal q of R is primary iff every zerodivisor in R/q is nilpotent. Neither the definition or primary ideal nor the characterization given in the above exercise is particularly enlightening, so one natural question is: which ideals are primary? (And, of course, another natural question is: what’s the significance of 158 PETE L. CLARK a primary ideal?) Here are some simple results which give some information on primary ideals, sufficient to determine all the primary ideals in some simple rings. Proposition 247. Let q be an ideal in a ring R. If r(q) = m is a maximal ideal, then q is primary. In particular, any power of a maximal ideal is primary. Proof. Since r(q) is the intersection of all prime ideals containing q, if this intersection is a maximal ideal m, then m is the unique prime ideal containing q and R/q is a local ring with nil(R/q) = J (R/q) = m/q. In such a ring an element is a zero-divisor iff it is a nonunit iff it is nilpotent, so q is primary. The “in particular” follows since by Proposition 122f), r(mn ) = r(m) = m. Proposition 248. If q is a primary ideal, then its radical r(q) is a prime ideal, the smallest prime ideal containing q. Proof. Let xy ∈ r(q), so that (xy )m = xm y m ∈ p for some m ∈ Z+ . If xm is in q then x ∈ r(q), so assume that xm is not in q. Then y m is a zero divisor in R/q, so by definition of primary there exists n ∈ Z+ such that (y m )n ∈ q, and then y ∈ r(q). The second statement holds for any ideal I whose radical is prime, since r(I ) is the intersection of all prime ideals containing I . A primary ideal is said to be p-primary if its radical is the prime ideal p. ∩n Lemma 249. If q1 , . . . , qn are p-primary ideals, then q = i=1 qi is also p-primary. Proof. Let x, y be elements of the ring R such that xy ∈ q and x ∈ R \ q. Then ∏n for all 1 ≤ i ≤ n, there exists ai ∈ Z+ such that y ai ∈ Ii , and then y i=1 ai ∈ q, so q is primary. Moreover, by Proposition 122b), n n n ∩ ∩ ∩ r(q) = r( qi ) = r(qi ) = q = q. i=1 i=1 i=1 Exercise 10.2: Give an example of primary ideals q, q′ such that q ∩ q′ is not primary. Proposition 250. If q is a primary ideal, the quotient ring R/q is connected. Proof. Indeed, a ring is disconnected if and only if it has an idempotent element e different from 0 or 1. Such an element is certainly not nilpotent en = e for all n – but is a zero-divisor, since e(1 − e) = e − e2 = 0. Exercise 10.3: Let k be a field, let R = k [x, y ] and put I = (xy ). Show that I is not primary but “nevertheless” R/I is connected. Example: We will find all primary ideals in the ring Z of integers. Evidently (0) is prime and hence primary. If q is any nonzero primary ideal, then its radical p = r(q) is a nonzero prime ideal, hence maximal. So, combining Propositions 247 and 248 we find that a nonzero ideal in Z is primary iff its radical is maximal. Moreover, for any prime power (pn ), r((pn )) = r((p)) = (p) is maximal – we use here the elementary and (we hope) familiar fact that if p is a prime number, (p) is a prime ideal (Euclid’s Lemma); such matters will be studied in more generality in §X.X on factorization – so (pn ) is a primary ideal. Conversely, if n is divisible by more than one prime power, then applying the Chinese Remainder Theorem, we get that Z/n is disconnected. COMMUTATIVE ALGEBRA 159 Exercise 10.4: a) Let R be an integral domain for which each nonzero ideal is a (finite, of course) product of maximal ideals. Use the above argument to show that an ideal q of R is primary iff it is a prime power. b) (For those who know something about PIDs) Deduce in particular that primary = prime power in any principal ideal domain. Remark: Consider the following property of an integral domain: (DD) Every ideal can be expressed as a product of prime ideals. This is a priori weaker than the hypothesis of Exericse X.Xa). Later we will devote quite a lot of attention to the class of domains satisfying (DD), the Dedekind domains. Among their many properties is that a Dedekind domain is (either a field or) a domain in which each nonzero prime ideal is maximal. Thus in fact the hypothesis of Exercise 10.4a) is equivalent to assuming that R is a Dedekind domain. Remark(ably): Another characterization theorem says that any Noetherian domain in which each primary ideal is a prime power is a Dedekind domain. In particular, any polynomial ring k [x1 , . . . , xn ] in 2 ≤ n < ∞ variables over a field admits primary ideals which are not prime powers. The following exercise gives an even simpler (and more explicit) example of a ring R and a primary ideal q of R which is not a prime power. √ Exercise 10.5: Let R = Z [t]/(t2 + 3) (or, equivalently, Z[ −3]). Let q = (2). a) Show that there is a unique ideal p2 with R/p2 = Z/2Z. Evidently p2 is maximal. b) Show that r(q) = p2 , and deduce that I is primary. c) Show that q is not a prime power, and indeed, cannot be expressed as a product of prime ideals.41 The ring R of Exercise 10.5 is a good one to keep in mind: it is simple enough to be easy to calculate with, but it already displays some interesting general phenomena. This is a Noetherian domain in which every nonzero prime ideal is maximal. It is therefore “close” to being a Dedekind domain but it does not satisfy one other property (“integral closure”) which will be studied later. It will turn out to be an immediate consequence of the main result of this section that, notwithstanding the fact that there are ideals which do not factor into a product of primes, nevertheless √ every proper ideal in R = Z[ −3] can be written as a product of primary ideals. This, finally, is some clue that the notion of a primary ideal is a fruitful concept. Having seen examples of a primary ideals which are not prime powers, what about the converse? Is it at any rate the case that any prime power is a primary ideal? We know that this is indeed the case for powers of a maximal ideal. However, the answer is again negative in general: 41Suggestion: show that R is a ring with finite quotients, and use properties of the norm function ||I || = #R/I . 160 PETE L. CLARK Example (Atiyah-MacDonald, p. 51): Let k be a field; put R = k [x, y, z ]/(xy − z 2 ). Denote by x, y , and z the images of x, y, z in R. Put p = ⟨x, z ⟩. Since R/p = k [x, y, z ]/(x, z, xy − z 2 ) = k [y ] is a domain, p is a prime ideal. Now consider the ideal p2 : we have xy = z 2 ∈ p2 , but x ∈ p2 and y ∈ p = r(p2 ), so p2 is not primary. / / 10.2. Primary decomposition, Lasker and Noether. Let R be a ring and I an ideal of R. A primary decomposition of I is an ∩n expression of I as a finite intersection of primary ideals, say I = i=1 qi . An ideal which admits at least one primary decomposition is said to be decomposable. This is not a piece of terminology that we will use often, but the reader should be aware of its existence. For any ring R, let us either agree that R itself admits the “empty” primary decomposition or that R has no primary decomposition (i.e., it doesn’t matter either way) and thereafter restrict our attention to proper ideals. It may not be too surprising that not every ideal in every ring admits a primary decomposition. Indeed, we will see later that if R is a ring for which (0) admits a primary decomposition, then the ring R has finitely many minimal primes. The first important result in this area was proved by Emanuel Lasker in 1905, roughly in the middle of his 27 year reign as world chess champion. Here it is. Theorem 251. (Lasker [Las05]) Let R be a polynomial ring in finitely many variables over a field. Then every proper ideal I of R admits a primary decomposition. Lasker’s proof of this theorem was a long and intricate calculation. As we will shortly see, a broader perspective yields considerably more for considerably less effort. In Lasker’s honor a ring R in which every proper ideal admits a primary decomposition is called a Laskerian ring. Exercise 10.6: If R is Laskerian and I is an ideal of R, then R/I is Laskerian. Combining Lasker’s theorem with this Exercise, we get that every finitely generated algebra over a field admits a primary decomposition. This result is of fundamental (indeed, foundational) importance in algebraic geometry. However, in 1921 Lasker’s triumph was undeniably trumped by Emmy Noether. Theorem 252. (Noether) Any proper ideal in a Noetherian ring admits a primary decomposition. In other words, a Noetherian ring is Laskerian. Therefore Lasker’s Theorem is an immediate consequence of Noether’s Theorem together with the Hilbert Basis Theorem, which we recall, was proved in 1888 and whose remarkably short and simple – but nonconstructive – proof engendered first controversy and later deep admiration. The same is true for Noether’s theorem: it is from this theorem, and the ridiculous simplicity of its proof, that Noetherian rings get their name. COMMUTATIVE ALGEBRA 161 Indeed, the result will practically prove itself once we introduce one further concept. An ideal I is irreducible if whenever I is written as an intersection of two ideals – i.e., I = J ∩ K – then I = J or I = K . Exercise 10.7: Let I be a proper ideal in a principal ideal domain R. TFAE: (i) I is primary. (ii) I is irreducible. (iii) I is a prime power: there exists a in R and n ∈ Z+ such that (a) is prime and I = (a)n = (an ). Dangerous bend: It follows therefore that if I = (x) is a principal ideal, then I being irreducible as an ideal is not the same as the element x being irreducible, even in very nice rings. For instance, it follows from Exercise X.X that an ideal I = (n) in Z is irreducible iff it is primary iff n = pr is a prime power, whereas of course n is irreducible as an element iff n = p is prime. For this reason, we are not thrilled with the term “irreducible.” We contemplated using the term “indecomposable” instead, but this ought to mean “not decomposable”, which is doesn’t. At last we have decided not to try to “correct” the terminology, as we will not be using it in the rest of these notes anyway. Proposition 253. a) A prime ideal is irreducible. b) An irreducible ideal in a Noetherian ring is primary. Proof. a) Let p be a prime ideal, and write p = I ∩ J . Since then p ⊃ IJ , by Proposition 118 we have p ⊃ I or p ⊃ J ; WLOG say p ⊃ I . Then p = I ∩ J ⊂ I ⊂ p, so that we must have I = p. b) By passage to the quotient, it is enough to assume that the 0 ideal is irreducible and show that it is primary. So, suppose that xy = 0 and x ̸= 0. Consider the chain of ideals ann(y ) ⊂ ann(y 2 ) ⊂ . . . ⊂ ann(y n ) ⊂ . . . . Since R is Noetherian, this chain stabilizes: there exists n such that ann(y n ) = ann(y n+k ) for all k . We claim that (x) ∩ (y n ) = 0. Indeed, if a ∈ (x) then ay = 0, and if a ∈ (y n ) then a = by n for some b ∈ R, hence by n+1 = ay = 0, so b ∈ ann(y n+1 ) = ann(y n ), hence a = by n = 0. Since the (0) ideal is irreducible, we must then have y n = 0, and this shows that (0) is primary. Proof of Noether’s Theorem : Let I be a proper ideal in the Noetherian ring R. We claim that I is a finite intersection of irreducible ideals; in view of Proposition 253 this gives the desired result. But the proof of this is a trivial application of Noetherian induction: suppose that the set of proper ideals which cannot be written as a finite intersection of irreducible ideals is nonempty, and choose a maximal element I . Then I is reducible, so we may write I = J ∩ K where each of J and K is strictly larger than I . But being strictly larger than I each of J and K can be written as a finite intersection of irreducible ideals, and hence so can I . Contradiction! 10.3. Irredundant primary decompositions. If an ideal can be expressed as a product of prime ideals, that product is in fact 162 PETE L. CLARK unique. We would like to have similar results for primary decomposition. Unfortunately such a uniqueness result is clearly impossible. Indeed, if I = q1 ∩ . . . ∩ qn is a primary decomposition of I and p is any prime containing I , then q1 ∩ . . . ∩ qn ∩ p is also a primary decomposition, and clearly a different one if p ̸= qi for any i. A proper ideal I may well be contained in infinitely many primes – e.g. by X.X this occurs with I = (0) for any Noetherian domain of dimension at least 2 – so there may well be infinitely many different primary decompositions. But of course throwing in extra primes is both frivolous and wasteful. The following definition formalizes the idea of a primary decomposition which is “frugal” in two reasonable ways. A primary decomposition is said to be irredundant42 (or minimal, or reduced) if both of the following properties hold: (IPD1) For all i ̸= j , r(qi ) ̸= r(qj ). ∩ (IPD2) For all i, qi does not contain j ̸=i qj . The following simple and important result tells us that if wastefulness succeeds, then so does frugality. Lemma 254. An ideal which admits a primary decompoistion admits an irredundant primary decomposition. Proof. By Lemma 249, we may replace any collection of primary ideals qi with a common radical with their intersection and still have a primary ideal, thus satisfying (IPD1). Then if (IPD2) is not satisfied, there is some qi which contains the intersection of all the other qj ’s, hence it can simply be removed to obtain a primary decomposition satisfying (IPD1) and with a smaller number of primary ideals. Clearly proceeding in this way, we eventually arrive at an irredundant primary decomposition. The question is now to what extent an irredundant primary decomposition is unique. The situation here is significantly better: although the primary decomposition is not in all cases unique, it turns out that there are some important quantities which are defined in terms of a primary decomposition and which can be shown to be independent of the choice of irredundant decomposition, i.e., are invariants of the ideal. Such uniqueness results are pursued in the next section. 10.4. Uniqueness properties of primary decomposition. Recall that for ideals I and J of a ring R, (I : J ) = {x ∈ R | xJ ⊂ I }, which is also an ideal of R. We abbreviate (I : (x)) to (I : x) and ((x) : J ) to (x : J ). Exercise: Show that for ideals I and J , I ⊂ (I : J ). Lemma 255. Let q be a p-primary ideal and x ∈ R. a) If x ∈ q then (q : x) = R. 42It is amusing to note that most dictionaries do not recognize “irredudant” as an English word, but mathematicians have been using it in this and other contexts for many years. COMMUTATIVE ALGEBRA 163 b) If x ∈ q then (q : x) is p-primary. / c) If x ∈ p then (q : x) = q. / Proof. a) If x ∈ q then 1(x) = x ⊂ q, so 1 ∈ (q : x). b) If y ∈ (q : x), then xy ∈ q; by assumption x ∈ q, so y n ∈ q for some n and thus y ∈ r(q) = p. So q ⊂ (q : x) ⊂ p; / taking radicals we get r((q : x)) = p. Moreover, if yz ∈ (q : x) with y ∈ (q : x), / then xyz = y (xz ) ∈ q, so (xz )n = xn z n ∈ q for some n, and xn ∈ q =⇒ (z m )n ∈ q / for some n ∈ Z+ , thus z mn ∈ q ⊂ (q : x). c) We have in all cases that q ⊂ (q : x). If x ∈ p = r(q) and y ∈ (q : x), then / xy ∈ q; since no power of x is q, we must have y ∈ q. ∩n Theorem 256. (First Uniqueness Theorem) Let I = i=1 qi be any irredundant primary decomposition of the ideal I . Let pi = r(qi ). Then the pi ’s are precisely the prime ideals of the form r((I : x)) as x ranges through elements of R. In particular, they are independent of the choice of irredundant primary decomposition. ∩ ∩ Proof. For x ∈ R we have (I : x) = ( i qi : x) = i (qi : x), so ∩ ∩ r((I : x)) = r((qi : x)) = pj x∈qj / i by Lemma 255. If r(I : x) is prime, then r(I : x) = pj for some∩ . Conversely, for each i, by irredundancy of the decomposition there j exists xi ∈ j ̸=i qj \ qi and then the Lemma implies r(I : xi ) = pi . ∩n Proposition 257. Let I = i=1 qi be a primary decomposition of an ideal I , with pi = r(qi ). Then any prime ideal p containing I contains pi for some i. ∩ Proof. If p ⊃ I = i qi , then ∩ ∩ p = r(p) ⊃ r(qi ) = pi , i i so by XX, p ⊃ pi for some i. Thus a minimal element of the set of associated primes of I is precisely a prime which is minimal over I (or equivalently, a minimal prime of R/I ). This has an important consequence: Corollary 258. Let I be a decomposable ideal in a ring R. Then the quotient ring R/I has only finitely many minimal primes. Thus a Laskerian ring – and, in particular, a Noetherian ring – has only finitely many minimal primes. In the next section we will give a rather different proof of the fact that a Noetherian ring has only finitely many minimal primes: we will use topological methods! Exercise 10.8: Show that an infinite Boolean ring is not Laskerian. It is therefore unambiguous to speak of a minimal associated prime of I . A synonym for a minimal associated prime is an isolated prime, whereas a nonisolated prime is often called an embedded prime.43 43It is, alas, not our intention to go deeply enough into theory to justify this terminology. 164 PETE L. CLARK Proposition 259. Let I ⊂ R be a decomposable ideal, I = primary decomposition, and pi = r(qi ). Then n ∪ pi = {x ∈ R : (I : x) ̸= I }. ∩n i=1 qi an irredundant i=1 In particular, if the zero ideal is decomposable, then the set of zero divisors of R is the union of the minimal primes. Proof. By passage to the quotient ring R/I , we may assume that I = 0. Let ∩r 0 = i=1 qi be a primary decomposition, with pi = r(qi ). For x ∈ R, ((0) : x) ̸= (0) iff x is a zero-divisor, so it suffices to show the last statement of the proposition, that the union of the minimal primes is the set of all zero-divisors. Let D be the set of all zero divisors, so from Exercise 3.X and the proof of Theorem 256 we have ∪ ∪∩ ∪ D = r (D ) = r((0 : x)) = pj ⊂ pj . 0̸=x 0̸=x x∈qj / j Conversely, by Theorem 256 each pi is of the form r((0 : x)) for some x ∈ R. Theorem 260. (Second Uniqueness Theorem) Let I be an ideal of R, and let n m ∩ ∩ qi = I = rj i=1 j =1 be two irredundant primary decompositions for an ideal I . By Theorem 256 we know that m = n and that there is a reordering r1 , . . . , rn of the rj ’s such that for 1 ≤ i ≤ n, r(qi ) = pi = r(ri ). Moreover, if pi is minimal, then qi = rj . In other words, the primary ideals corresponding to the minimal primes are independent of the primary decomposition. We will use the technique of localization to prove this result, so first we need some preliminaries on the effect of localization on a primary decomposition. Proposition 261. Let R be a ring, S ⊂ R a multiplicatively closed set, and q be a p-primary ideal. Write ι : R → S −1 R for the localization map. a) If S ∩ p ̸= ∅, then ι∗ (q) = S −1 R. b) If S ∩ p = ∅, then ι∗ (q) is ι∗ (p)-primary, and ι∗ (ι∗ (q)) = q. Proof. a) If x ∈ S ∩ p, then for some n ∈ Z+ , xn ∈ S ∩ q, so ι∗ (q) contains a unit of S −1 R and is therefore S −1 R. Part b) follows immediately from Proposition 165 and Proposition 167a). ∩n Proposition 262. Let S ⊂ R be a multiplicatively closed set, and let I = i=1 qi be an irredundant primary decomposition of an ideal I . Put pi = r(qi ) and suppose that the numbering is such that S ∩ pi = ∅ for i ≤ m and S ∩ pi ̸= ∅ for i > m. Then: ι∗ (I ) = m ∩ ι∗ (qi ), i=1 ι∗ ι∗ (I ) = m ∩ qi , i=1 and both of these are irrendundant primary decompositions. COMMUTATIVE ALGEBRA 165 Exercise 10.9: Prove Proposition 262. Proof of Theorem 260: let pi be a minimal associated prime, and put S = R \ pi . Certainly S is a multiplcatively closed set, and moreover by minimality pi is the unique associated prime which is disjoint from S . Applying Proposition 262 to both primary decompositions gives qi = ι∗ ι∗ (I ) = ri . 10.5. Applications in dimension zero. We now give the proof of the uniqueness portion of Theorem 217. Let m1 , . . . , mn be the distinct maximal ideals of the Artinian ring R. As in the proof of Theorem ∏n 217a) there exists k ∈ Z+ such that i=1 mk = ∩n mk = 0. For each i, the radical i i=1 i r(mk ) is the maximal ideal mi , so by Proposition 247 each mk is an mi -primary i i ∩n ideal. Thus 0 = i=1 mk is a primary decomposition of the zero ideal which is i moreover immediately seen to be irredundant. Since all the primes mi are maximal, the desired uniqueness statement of Theorem 217b) follows from the Second Uniqueness Theorem (Theorem 260) for primary decompositions. 10.6. Applications in dimension one. Let R be a one-dimensional Noetherian domain, and∩I a nonzero ideal. Then n by Theorem 252, I has a primary decomposition: I = i=1 qi , where pi = r(qi ) ⊃ qi ⊃ I is a nonzero prime ideal. But therefore each pi is maximal, so that the pi ’s are pairwise comaximal. By Proposition 124, so too are the qi ’s, so the Chinese Remainder Theorem applies to give I= n ∩ qi = i=1 n ∏ qi , i=1 and R/I ∼ = n ∏ R/qi . i=1 Thus in this case we can decompose any proper ideal as a finite product of primary ideals and not just a finite intersection. Moreover, for I ̸= 0, all the associated primes are minimal over I , so the Uniqueness Theorems (Theorems 256 and 260) simply assert that the ideals qi are unique. This observation will be very useful in our later study of ideal theory in one dimensional Noetherian domains. 11. Affine k -algebras and the Nullstellensatz Let k be a field. By an affine algebra over k we simply mean a finitely generated k -algebra. Of all the various and sundry classes of commutative rings we have met and will meet later in these notes, affine algebras are probably the most important and most heavily studied, because of their connection to algebraic geometry. 166 PETE L. CLARK 11.1. Zariski’s Lemma. In 1947 the great algebraic geometer Oscar Zariski published a short note [Zar47] proving the following result. Theorem 263. (Zariski’s Lemma) Let k be a field, A a finitely generated k -algebra and m a maximal ideal of A. Then A/m is a finite degree field extension of k . Exercise X.X: Show that the following is an equivalent restatement of Zariski’s Lemma: let K/k be a field extension such that K is finitely generated as a k algebra. Then K/k is an algebraic field extension. Notwithstanding its innocuous initial appearance, Zariski’s Lemma is a useful result for affine algebras over an aribtrary field which, in the special case that k is algebraically closed, carries all of the content of Hilbert’s Nullstellensatz, the main theorem of this section. So how do we prove Zariski’s Lemma? Oh, let us count the ways! The literature contains many interesting proofs, employing an impressively wide range of ideas and prior technology. We will in fact give several different proofs during the course of these notes. Of course some pride of place goes to the first proof that we give, so after much thought (and after changing our mind at least once!) we have decided on the following. Proof of Zariski’s Lemma via the Artin-Tate Lemma: As in Exercise X.X, it suffices to prove the following: let K/k be a field extension which is finitely generated as a k -algebra. Then K/k is algebraic. Well, suppose otherwise: let x1 , . . . , xn be a transcendence basis for K/k (where n ≥ 1 since K/k is transcendental), put k (x) = k (x1 , . . . , xn ) and consider the tower of rings (16) k ⊂ k (x) ⊂ K. To be sure, we recall the definition of a transcendence basis: the elements xi are algebraically equivalent over k and K/k (x) is algebraic. But since K is a finitely generated k -algebra, it is certainly a finitely generated k (x)-algebra and thus K/k (x) is a finite degree field extension. Thus the Artin-Tate Lemma applies to (16): we conclude that k (x)/k is a finitely generated k -algebra. But this is absurd. It implies the much weaker statement that k (x) = k (x1 , . . . , xn−1 )(xn ) is finitely generated as a k (x1 , . . . , xn−1 )[xn ]-algebra, or weaker yet, that there exists some field F such that F (t) is finitely generated as an F [t]-algebra: i.e., there exist finitely many rai (t) tional functions {ri (t) = pi (t) }N such that every rational function is a polynomial i=1 q in the ri ’s with k -coefficients. But F [t] is a PID with infinitely many nonassociate nonzero prime elements q (e.g. adapt Euclid’s argument of the infinitude of the primes), so we may choose a nonzero prime element q which does not divide qi (t) for any i. It is then clear that 1 cannot be a polynomial in the ri (t)’s: for instance, q evaluation at a root of q in F leads to a contradiction. Remark: The phenomenon encountered in the endgame of the preceding proof COMMUTATIVE ALGEBRA 167 will be studied in great detail in §12. What we are actually showing is that for any field F , the polynomial ring F [t] is not a Goldman domain, and indeed this is closely related to the fact that Spec F [t] is infinite. More on this later! 11.2. Hilbert’s Nullstellensatz. Let k be a field, let Rn = k [t1 , . . . , tn ], and write An for k n . We wish to introduce an antitone Galois connection (V, I ) between subsets of Rn and subsets of An . Namely: For S ⊂ An , we put I (S ) = {f ∈ Rn | ∀x ∈ S, f (x) = 0}. In other words, I (S ) is the set of polynomials which vanish at every element of S . Conversely, for J ⊂ Rn , we put V (J ) = {x ∈ An | ∀f ∈ J, f (x) = 0}. This is nothing else than the Galois relation associated to the relation f (x) = 0 on the Cartesian product Rn × An . As usual, we would like to say something about the induced closure operators on Rn and An . First, for any subset S of An , I (S ) is not just a subset but an ideal of Rn . In fact I (S ) is a radical ideal: indeed, if f n ∈ I (S ) then f n vanishes on every point of S , so f vanishes at every point of S . This little bit of structure pulled from thin air will quicken the heart of any Bourbakiste. But beyond the formalism, the key question is: exactly which sets are closed? Without knowing this, we haven’t proved the Nullstellensatz any more than the analogous formalities between sets and groups of automorphisms prove the Galois correspondence for Galois field extensions. Indeed, an ideal I is radical if f n ∈ I implies f ∈ I . But if f n vanishes identically on S , then so does f . The closed subsets of An are closed under arbitrary intersections (including the “empty intersection”: An = V ((0)) and under finite unions (including the “empty union”: ∅ = V ({1}) = V (Rn ), and therefore form the closed sets for a unique topology on An , the Zariski topology. Exercise: a) Prove these facts. b) Show that the Zariski topology on An coincides with the topology it inherits as /k a subset of Ank . / c) Show that the Zariski topology is T1 : i.e., singleton subsets are closed. d) Show that when n = 1, the Zariski topology is the coarsest T1 topology on k : namely, the topology in which a proper subset is closed iff it is finite. e) For any n ≥ 1, show that the Zariski topology on k n is discrete iff k is finite. f) For any infinite field and m, n ≥ 1, show that the Zariski topology on k m+n is strictly finer than the product of the Zariski topologies on k m and k n . 168 PETE L. CLARK Remark: It is often lamented that the Zariski topology (especially when k = C) is distressingly “coarse”. It is true that it is much coarser than the “analytic topology” on k n when k is a topological field (i.e., the product topology from the topology on k ). But in fact from an algebraic perspective the Zariski topology is if anything too fine: we will see why later on when we extend the topology to all prime ideals on an affine algebra. But also the fact that the Zariski topology on a Fn is discrete q creates many geometric problems. Exercise: Let k be a field, n ∈ Z+ as above. Explicitly compute the ideal I (k n ), i.e., the set of all polynomials which vanish at every point of k n . Do we necessarily have I (k n ) = {0}? (Hint: No, but we can figure out exactly for which fields this occurs.) Lemma 264. For a = (a1 , . . . , an ) ∈ k n , put ma = ⟨x1 − a1 , . . . , xn − an ⟩. Then: a) We have Rn /ma = k . In particular ma is maximal. b) ma = I ({a}) is the ideal of all functions vanishing at a. c) The assignment a → ma is a bijection from k n to the set of all maximal ideals m of Rn such that Rn /m = k . Proof. Part a) is obvious (but important). b) Certainly each xi − ai vanishes at a, so ma ⊂ I ({a}). But by part a) ma is a maximal ideal, whereas 1 ∈ I ({a}), so we must have ma = I ({a}). / c) The mapping a → ma is certainly an injection from k n to the set of maximal ideals with residue field k . Conversely, let m be an ideal of Rn such that R/m = k . For 1 ≤ i ≤ n let ai be the image of xi in R/m = k . Then m ⊃ ma so again we must have equality. We now pause for a very important definition. A ring R is a Jacobson ring if it is “sufficiently many maximal ideals”: more precisely, such that every prime ideal p of R is the intersection of the maximal ideals that contain it. Exercise X.X: a) Show that a ring R is a Jacobson ring iff for every ideal I , the intersection of all maximal ideals containing I is rad(I ). b) Show that every homomorphic image of a Jacobson ring is Jacobson. Proposition 265. (Rabinowitsch Trick [Ra30]) Let k be any field and n ∈ Z+ . a) The ring Rn = k [x1 , . . . , xn ] is a Jacobson ring. b) It follows that any affine algebra is a Jacobson ring. Proof. a) It is sufficient to show that for each prime ideal p of Rn and a ∈ R \ p, there exists a maximal ideal m containing p and not containing a. 1 To show this, put Ra := R[ a ], and let pa = pRa be the pushed forward ideal. Since p does not meet the multiplicative set generated by a, pa is still prime in Ra . Let ma be any maximal ideal of Ra conaining pa , and let m = ma ∩ R be its contraction to R: a priori, this is a prime ideal. There is an induced k -algebra embedding R/m → Ra /ma . But Ra is still a finitely generated algebra so by Zariski’s Lemma (Theorem 263) Ra /ma is finite dimensional as a k -vector space, hence so is the subspace R/m. Thus the domain R/m must be a field, a special case of our coming work on integral extensions. Concretely, let x ∈ (R/m)• and write out a linear dependence relation of minimal degree among the powers of x: xn + cn−1 xn−1 + . . . + c1 x + c0 , ci ∈ k, c0 ̸= 0. COMMUTATIVE ALGEBRA 169 Thus ( ) −1 x xn−1 + cn−1 xn−2 + . . . + c1 = , c0 so x is invertible. Thus m is the desired maximal ideal. b) This follows immediately from Exercise X.X. Remark: It seems that “J.L. Rabinowitch”, the author of [Ra30], is the same person as George Yuri Rainich, a distinguished Russo-American mathematician of the early twentieth century. Here is one last fact we can prove before imposing the hypothesis that k is algebraically closed. Proposition 266. Let k be any field and J an ideal of k [x] = k [x1 , . . . , xn ]. Then: a) V (J ) = V (rad J ). b) For any subset S ⊂ k n , I (S ) is a radical ideal. v) I (V (J )) is a radical ideal containing rad J . Proof. The underlying mechanism here is the following truly basic observation: for f ∈ k [x], P ∈ k n and m ∈ Z+ , we have f (P ) = 0 ⇐⇒ f m (P ) = 0. a) Since J ⊂ rad J we have V (J ) ⊃ V (rad J ). Conversely, let P ∈ V (J ) and f ∈ rad J . Then there exists m ∈ Z+ such that f m ∈ J , so f m (P ) = 0 and thus f (P ) = 0. It follows that P ∈ V (rad J ). b) Similarly, for any f ∈ k [x] and m ∈ Z+ , if f m ∈ I (S ), then for all P ∈ S , f m (P ) = 0. But this implies f (P ) = 0 for all P ∈ S and thus f ∈ I (S ). c) This follows immediately from parts a) and b) and the tautological fact that for any ideal J of k [x], I (V (J )) ⊃ J . Finally we specialize to the case in which the field k is algebraically closed. We have done almost all the work necessary to establish the following fundamental result. Theorem 267. (Hilbert’s Nullstellensatz) Let k be an algebraically closed field, let k [x] = k [x1 , . . . , xn ]. Then: a) I induces a bijective correspondence between the singleton sets of k n and the maximal ideals: a ∈ k n → ma = ⟨x1 − a1 , . . . , xn − an ⟩. b) For any Zariski-closed subset S ⊂ k n , V (I (S )) = S . c) For any ideal J of Rn , I (V (J )) = rad(J ). Thus there is an inclusion-reversing, bijective correspondence between Zariski-closed subsets of k n and radical ideals of k [x]. Proof. a) Let m be a maximal ideal of k [x]. By Theorem 263, the residue field k [x]/m is a finite degree extension of k . Since k is algebraically closed, this forces k [x]/m = k , and now Lemma 264 applies. b) There is no content here: it is part of the formalism of Galois connections. c) By Proposition 266, it is no loss of generality to assume that J is a radical ideal. Further, by Proposition 265, k [x] is a Jacobson ring, so any radical ideal J is the intersection of the maximal ideals m containing it. This is true over any field k . But combining with part a), we get that J is an intersection of maximal ideals of the form ma for certain points a ∈ k n . Since ma = I ({a}), J ⊂ ma iff every element of J vanishes at a, in other words iff a ∈ V (J ). Thus J is equal to the set of all polynomials f ∈ Rn which vanish at every point of V (J ): J = I (V (J ))! 170 PETE L. CLARK Exercise: Let k be any field. Show that if either part a) or part c) of Theorem 267 holds for the rings k [x1 , . . . , xn ], then k is algebraically closed. (Hint: in fact both parts fail for each n ∈ Z+ , including n = 1.) Exercise: Show that Zariski’s Lemma in the case that k is algebraically closed is equivalent to the following statement: let J = ⟨f1 , . . . , fm ⟩ be an ideal in k [x1 , . . . , xn ]. Then either there exists a simultaneous zero a of f1 , . . . , fm or there exist polynomials g1 , . . . , gm such that g1 f1 + . . . + gm fm = 1. 11.3. Other Nullstellens¨tze. a 11.3.1. The Real Nullstellensatz. Recall that a field k is formally real if it is not possible to express −1 as a sum of (any finite number of) squares in k . Exercise: Let k be a formally real field. a) Show that k is not algebraically closed. b) Show that any subfield of k is formally real. A field k is real closed if it is formally real and admits no proper formally real algebraic extension. So for example R is real closed and Q is formally real but not real closed. As we saw, even the weak Nullstellensatz fails for polynomial rings over any nonalgebraically closed field k . However, when k is formally real one can find counterexamples to the Nullstellensatz of a particular form, and when k is real closed one can show that these counterexamples are in a certain precise sense the only ones, leading to an identification of the closure operation J → I (V (J )) in this case. In any commutative ring R, an ideal I is real if for all n ∈ Z+ and x1 , . . . , xn ∈ R, x2 + . . . + x2 ∈ I implies x1 , . . . , xn ∈ I . n 1 Exercise: Show that a prime ideal p of R is real iff the fraction field of R/p is a formally real field. Exercise: Show that any real ideal is a radical ideal. So what? The following result gives the connection to the Nullstellensatz. Proposition 268. Let k be a formally real field and k [x] = k [x1 , . . . , xn ]. For any ideal J of k [x], its closure J = I (V (J )) is a real ideal. 2 2 Proof. Let f1 , . . . , fm ∈ k [x] be such that f1 + . . . + fm ∈ J . Then for any P ∈ 2 2 V (J ), we have f1 (P ) + . . . + fm (P ) = 0. Since k is formally real, this implies f1 (P ) = . . . = fm (P ), and thus f1 , . . . , fm ∈ I (V (J )) = J . 12. Goldman domains and Hilbert-Jacobson rings COMMUTATIVE ALGEBRA 171 12.1. Goldman domains. Lemma 269. Let R be a domain with fraction field K . TFAE: (i) K is finitely generated as an R-algebra. (ii) There exists f ∈ K such that K = R[f ]. Proof. Of course (ii) =⇒ (i). Conversely, if K = R[f1 , . . . , fn ], then write fi = 1 and then K = R[ q1 ···qn ]. pi qi , A ring satisfying the conditions of Lemma 269 will be called a Goldman domain. Exercise: Show that an overring44 of a Goldman domain is a Goldman domain. Lemma 270. Let R be a domain with fraction field K , and 0 ̸= x ∈ R. TFAE: (i) Any nonzero prime ideal of R contains x. (ii) Any nonzero ideal contains a power of x. (iii) K = R[x−1 ]. Proof. (i) =⇒ (ii): let I be a nonzero ideal. If I is disjoint from {xn }, then by Multiplicative Avoidance (117), I can be extended to a prime ideal disjoint from {xn }, contradicting (i). (ii) =⇒ (iii): Let 0 ̸= y ∈ R. By (ii), we have (y ) contains some power of x, say xk = yz . But this implies that y is a unit in R[x−1 ]. (iii) =⇒ (i): The prime ideals that are killed in the localization map R → R[x−1 ] are precisely those which meet the multiplicatively closed set {xk }, i.e., contain x. Corollary 271. For an integral domain R, TFAE: (i) R is a Goldman domain. (ii) The intersection of all nonzero prime ideals of R is nonzero. Exercise X.X: Prove Corollary 271. Easy examples of Goldman domains: a field, k [[t]], Z(p) . In fact we have developed enough technology to give a remarkably clean characterization of Noetherian Goldman domains. Theorem 272. Let R be an integral domain. a) If R has only finitely many primes, then R is a Goldman domain. b) If R is a Noetherian Goldman domain, then R has finitely many primes. c) A Noetherian Goldman domain is either a field or a one-dimensional domain. Proof. It is harmless to assume rhroughout that R is not a field, and we do so. a) Suppose that R has only finitely many primes, and let p1 , . . . , pn be the nonzero prime ideals of R. For 1 ≤ i ≤ n, let 0 ̸= xi ∈ pi , and put x = x1 · · · xn . Then the multiplicative set S generated by x meets every nonzero prime of R, so 1 that S −1 R has only the zero ideal. In other words, R[ x ] is the fraction field of R, so R is a Goldman domain. (Alternately, this is an almost immediate consequence of Corollary 271.) 1 b) Similarly, for a Goldman domain R we can write K = R[ x ] for x ∈ R and then every nonzero prime of R contains x. Suppose first that (x) itself is prime, necessarily of height one by the Hauptidealsatz, hence if R has any primes other than (0) and (x) – especially, if it has infinitely many primes – then it has a height 44An overring of a domain R is a ring intermediate between R and its fraction field K . 172 PETE L. CLARK two prime q. But by Corollary 226 a Noetherian ring cannot have a height two prime unless it has infinitely many height one primes, a contradiction. So we may assume that (x) is not prime, and then the minimal primes of the Noetherian ring R/(x) are finite in number – say p1 , . . . , pn – and correspond to the primes of R which are minimal over x, so again by the Hauptidealsatz they all have height one. Similarly, if R has infinitely many primes there would be, for at least one i (say i = 1), a height two prime q ⊃ p1 . But then by Corollary 226 the “interval” (0, q) is infinite. Each element of this set is a height one prime ideal containing (x), i.e., is one of the pi ’s, a contradiction. Finally, part c) follows by once again applying Corollary 226: a Noetherian ring of dimension at least two must have infinitely many primes. Remark: a non-Noetherian Goldman domain can indeed have infinitely many primes and/or primes of arbitrarily large height. Proposition 273. Let R be an integral domain. Then the polynomial ring R[t] is not a Goldman domain. Proof. Let K be the fraction field of R. If R[t] is a Goldman domain, then by Exercise XX, so is its overring K [t]. But the ring K [t] is a Noetherian domain with infinitely many primes – for instance, Euclid’s proof of the infinitude of primes in Z carries over verbatim to K [t] – so Theorem 272 applies to show that K [t] is not a Goldman domain. Proposition 274. Let R be a domain, and T ⊃ R an extension domain which is algebraic and finitely generated as an R-algebra. Then R is a Goldman domain iff T is a Goldman domain. Proof. Let K and L be the fraction fields of R and T , respectively. Suppose first 1 1 that R is a Goldman domain: say K = R[ u ]. Then T [ u ] is algebraic over the 1 field K , so is a field, hence we have L = T [ u ]. Conversely, suppose that T is 1 a Goldman domain: say L = T [ v ]; also write T = R[x1 , . . . , xk ]. The elements v −1 , x1 , . . . , xk are algebraic over R hence satisfy polynomial equations with coefficients in R. Let a be the leading coefficient of a polynomial equation for v −1 and b1 , . . . , bk be the leading coefficients of polynomial equations for x1 , . . . , xk . Let R1 := R[a−1 , b−1 , . . . , b−1 ]. Now L is generated over R1 by x1 , . . . , xk , v −1 , all of 1 k which are integral over R1 . Therefore L is integral over R1 . Since L is a field, it follows that R1 is a field, necessarily equal to K , and this shows that R is a Goldman domain. Corollary 275. Let R ⊂ S be an inclusion of domains, with R a Goldman domain. Suppose that u ∈ S is such that R[u] is a Goldman domain. Then u is algebraic over R, and R is a Goldman domain. Theorem 276. For an integral domain R, TFAE: (i) R is a Goldman domain. (ii) There exists a maximal ideal m of R[t] such that m ∩ R = (0). 1 Proof. (i) =⇒ (ii): We may assume WLOG that R is not a field. Write K = R[ u ]. 1 Define a homomorphism φ : R[t] → K by sending t → u . Evidently φ is surjective, so its kernel m is a maximal ideal, and clearly we have m ∩ R = 0. (ii) =⇒ (i): Suppose m is a maximal ideal of R[t] such that m ∩ R = (0). Let COMMUTATIVE ALGEBRA 173 v be the image of t under the natural homomorphism R[t] → R[t]/m. Then R[v ] is a field, so by Corollary 275, R is a Goldman domain. We define a prime ideal p of a ring R to be a Goldman ideal if R/p is a Goldman domain. Write G Spec R for the set of all Goldman ideals. Thus a Goldman ideal is more general than a maximal ideal but much more special than a prime ideal. Proposition 277. Let R be a ring and I an ideal of R. a) The nilradical of R is the intersection of all Goldman ideals of R. b) The radical of I is the intersection of all Goldman ideals containing I . ∩ ∩ Proof. a) We know that N = p∈Spec R p, so certainly N ⊂ p∈G Spec R p. Conversely, suppose x ∈ R \ N . The ideal (0) is then disjoint from the multiplicative set S = {xn }. By multiplicative avoidance, we can extend (0) to an ideal p maximal with respect to disjointness from S . We showed earlier that p is prime; we now claim that it is a Goldman ideal. Indeed, let x denote the image of x in R = R/p. By maximality of p, every nonzero prime of R contains x. By Lemma 270, this implies R[x−1 ] is a field, thus R is a Goldman domain, and therefore p is a Goldman ideal which does not contain x. Part b) follows by correspondence, as usual. The following result may seem completely abstruse at the moment, but soon enough it will turn out to be the key: Corollary 278. An ideal I in a ring R is a Goldman ideal iff it is the contraction of a maximal ideal in the polynomial ring R[t]. Proof. This follows from Theorem 276 by applying the correspondence principle to the quotient ring R/I . Theorem 279. a) Let M be a maximal ideal in R[t], and suppose that its contraction m = M ∩ R is maximal in R. Then M can be generated by m and by one additional element f , which can be taken to be a monic polynomial which maps modulo m to an irreducible polynomial in R/m[t]. b) If, moreover, we suppose that R/m is algebraically closed, then M = ⟨m, t − a⟩ for some a ∈ R. Proof. a) Since M contains m, by correspondence M may be viewed as a maximal ideal of R[t]/mR[t] ∼ (R/m)[t], a PID, so corresponds to an irreducible polynomial = f ∈ R/m[t]. If f is any lift of f to R[t], then M = ⟨m, f ⟩. Part b) follows immediately from the observation that an irreducible univariate polynomial over an algebraically closed field is linear. The following more elementary result covers the other extreme. Theorem 280. Let R be a domain, with fraction field K . Let ι : R[t] → K [t] be the natural inclusion. Then ι∗ induces a bijection between the prime ideals P of R[t] such that P ∩ R = {0} and the prime ideals of K [t]. Proof. Let S = R \{0}. The key observation is that S −1 R[t] = K [t]. Recall (Proposition 167) that in any localization map R → S −1 R, the prime ideals which push forward to the unit ideal are precisely those which meet S , whereas the localization map restricted to all other prime ideals is a bijection onto the set of prime ideals of S −1 R. Applying that in this case gives the desired result immediately! 174 PETE L. CLARK 12.2. Hilbert rings. To put Theorem 279 to good use, we need to have a class of rings for which the contraction of a maximal ideal from a polynomial ring is again a maximal ideal. It turns out that the following is the right class of rings: Definition: A Hilbert ring is a ring in which every Goldman ideal is maximal. Proposition 281. Any quotient ring of a Hilbert ring is a Hilbert ring. Proof. This follows immediately from the correspondence between ideals of R/I and ideals of R containing I . A direct consequence of the definition and Proposition 277 is the following: Proposition 282. Let I be an ideal in a Hilbert ring R. Then the intersection ∩ m sup I m of all maximal ideals m containing I is rad(I ). Examples: Any zero dimensional ring is a Hilbert ring: in this case all prime ideals are maximal. Especially, a field is a Hilbert ring, as is any Artinian ring, or any Boolean ring. Exercise: a) Let R be a one-dimensional Noetherian domain. TFAE: (i) R is a Hilbert ring. (ii) The Jacobson radical of R is 0. (iii) R has infinitely many prime ideals. (iv) R is not a Goldman domain. b) Deduce that the ring Z of integers is a Hilbert domain. Theorem 283. Let R be a Hilbert ring, and S a finitely generated R-algebra. Then: a) S is also a Hilbert ring. b) For every maximal ideal P of S , p := P ∩ R is a maximal ideal of R. c) The degree [S/P : R/p] is finite. Proof. a) It suffices to show that R is a Hilbert ring iff R[t] is a Hilbert ring, for then, if R is a Hilbert ring, by induction any polynomial ring R[t1 , . . . , tn ] is a Hilbert ring, and any finitely generated R-algebra is a quotient of R[t1 , . . . , tn ] and thus a Hilbert ring. Note also that since R is a homomorphic image of R[t], if R[t] is a Hilbert domain, then so also is R. So suppose that R is a Hilbert ring, and let q be a Goldman ideal in R[t]; we must show that q is maximal. Put p = q ∩ R. As above, we can easily reduce to the case p = 0, so that in particular R is a domain. Let a be the image of t in the natural homomorphism R[t] → R[t]/q. Then R[a] is a Goldman domain. By Corollary 275, a is algebraic over R and R is a Goldman domain. But since we assumed that R was a Hilbert ring, this means that R is a field, and thus R[a] = R[t]/q is a field, so q is maximal. b) We may write S = R[t1 , . . . , tn ]/I . A maximal ideal m of S is just a maximal ideal of R[t1 , . . . , tn ] containing I . By Corollary 278, the contraction m′ of m to R[t1 , . . . , tn−1 ] is a Goldman ideal of the Hilbert ring R[t1 , . . . , tn−1 ], so is therefore maximal. Moreover, by Theorem 279, m is generated by m′ and an irreducible polynomial in R/m′ [t], so that the residual extension R[t1 , . . . , tn ]/m has finite degree over R[t1 , . . . , tn−1 /m′ . Again, induction gives the full result. COMMUTATIVE ALGEBRA 175 Applying Theorem 283c) in the case R = k is a field, we deduce our second proof of Zariski’s Lemma (Lemma 263). Moreover, we deduce a strong form of the Dimension Theorem for Noetherian Hilbert rings. Theorem 284. Let R be a Noetherian Hilbert ring. Then dim(R[t]) = dim R + 1. Proof. Let 0 = p0 p1 ... pd be a chain of prime ideals in R. Then, with ι : R → R[t] the natural inclusion, ι∗ p0 ... ι∗ pn ⟨ι∗ (pn ), t⟩ is a chain of prime ideals of R[t] of length d + 1, hence for any ring R we have dim R[t] ≥ dim R +1. Conversely, it suffices to show that the height of any maximal ideal P of R[t] is at most d + 1. For this, put p = P ∩ R. By Theorem 283, p is maximal in R, so Theorem 279 tells us that there exists f ∈ R[t] such that P = ⟨ι∗ p, f ⟩. Applying Krull’s Hauptidealsatz in the quotient ring R[t]/ι∗ p, we get that the height of P is at most one more than the height of p. Corollary 285. Let k be a field, and put R = k [t1 , . . . , tn ]. a) Then every maximal ideal of R has height n and can be generated by n elements (and no fewer, by Theorem 228). b) In particular, dim R = n. Exercise: Prove Corollary 285. 12.3. Jacobson Rings. We define a Jacobson ring to be a ring which satisfies the conclusion of Proposition 282, i.e., a ring R in which for every ideal I , the radical of I is the intersection of all the maximal ideals containing I . Exercise: Show that for a ring R, TFAE: (i) R is a Jacobson ring. (ii) Every prime ideal of R is an intersection of maximal ideals. (iii) Every radical ideal of R is an intersection of maximal ideals. Exercise: Show that any quotient of a Jacobson ring is a Jacobson ring. 12.4. Hilbert-Jacobson Rings. Proposition 286. Suppose R is both a Goldman domain and a Jacobson ring. Then R is a field. Proof. Let K be the fraction field of R, and suppose for a conradiction that R ̸= K . Then there exists a nonzero nonunit f ∈ R such that K is the localization of R at the multiplicative subset S = {f, f 2 , . . .}. Let m be a maximal ideal of R. Since R is not a field, m is not zero, and thus the pushforward of R to S −1 R is the unit ideal. By Prop. X.X, this implies that m meets S . Since m is prime, we conclude f ∈ m. It follows that the Jacobson radical of R contains f is accordingly nonzero. On the other hand R, being a domain, has zero nilradical. Thus R is not Jacobson. 176 PETE L. CLARK Our previous work now immediately gives the following result: Theorem 287. For a commutative ring R, TFAE: (i) R is a Hilbert ring. (ii) R is a Jacobson ring. (iii) For all maximal ideals m of R[t], m ∩ R is a maximal ideal of R. Proof. (i) =⇒ (ii) by Proposition 282. (ii) =⇒ (i): Suppose R is Jacobson and p is a Goldman ideal of R. By Exercise X.X, the Goldman domain R/p is also Jacobson, hence by Proposition 286 R/p is a field, so p is maximal. (ii) =⇒ (iii) is Theorem 283b). (iii) =⇒ (i): Suppose R is a ring such that every maximal ideal of R[t] contracts to a maximal ideal of R, and let p be a Goldman ideal of R. By Corollary 278, p is the contraction of a maximal ideal of R[t], hence by assumption p is maximal. In the sequel we will use the consolidated terminology Hilbert-Jacobson ring for a ring satisfying the equivalent conditions of Theorem 287. 13. The spectrum of a ring 13.1. The Zariski spectrum. For a commutative ring R, we denote the set of all prime ideals of R by Spec R. Moreover, we refer to Spec R as the Zariski spectrum – or prime spectrum – of the ring R. It is important to notice that Spec R comes with additional structure. First, it has a natural partial ordering, in which the maximal elements are the maximal ideals, and the minimal elements are (by definition) the minimal primes. Also, as O. Zariski first observed, Spec R can be endowed with a topology. To see this, for any ideal I of R, put V (I ) = {p ∈ Spec R | p ⊃ I }. Proposition 288. For ideals I and J of R, we have V (I ) = V (J ) iff rad I = rad J . Proof. For any ideal I and any prime ideal p, p ⊃ I iff p ⊃ rad I , and therefore V (I ) = V (rad I ). Conversely, if V (I ) = V (J ), then the set of prime ideals containing I is the same as the set of prime ideals containing J . So ∩ ∩ rad I = p= p = rad J. p⊃I p⊃J Exercise X.X: Show that I ⊂ J iff V (I ) ⊃ V (J ). Now we claim that the family of subsets V (I ) of Spec R has the following properties: (ZT1) ∅ = V (R), Spec R = V ((0)). ∩ (ZT2) If {Ii } is any collection of ideals of R, then i V (Ii ) = V (⟨Ii ⟩). ∩ ∪n n (ZT3) If I1 , . . . , In are ideals of R, then i=1 V (Ii ) = V (I1 · · · In ) = V ( i=1 Ii ). COMMUTATIVE ALGEBRA 177 ∩ (ZT1) is obvious. As for (ZT2), let p be a prime ideal of R. Then p ∈ i V (Ii ) for all i iff p ⊃ Ii for all i iff p contains the ideal generated by all Ii . As for (ZT3), p contains a product of ideals iff it contains one of the ideals of the product. Therefore there is a unique topology on Spec R in which the closed sets are precisely those of the form V (I ). This is called the Zariski topology. It is of course natural to ask for a characterization of the open sets. Recall that a base for the open sets of a topology is a collection {Bi } of open sets such that: (BT1) for any point x ∈ Bi ∩ Bj , there exists a k such that x ∈ Bk ⊂ Bi ∩ Bj ; (BT2) every open set is a union of the Bi ’s contained in it. For f ∈ R, we define U (f ) := Spec R \ V ((f )). In other words, U (f ) is the collection of all prime ideals which do not contain the element f . For f, g ∈ R, U (f ) ∩ U (g ) is the set of prime ideals p containing neither f nor g ; since p is prime, this is equivalent to p not containing f g , thus U (f ) ∩ U (g ) = U (f g ), which is a stronger property than (BT1). Moreover, any open set U is of ∩ form the Spec R \ V (I ). Each ideal I is the union of all of its elements fi , so V (I ) = i V (fi ), so that ∩ ∪ ∪ U = Spec R \ V (I ) = Spec R \ V (fi ) = (Spec R \ V (fi )) = U (fi ). i i i Proposition 289. Let R be any ring, and consider the canonical homomorphism f : R → Rred = R/ nil(A). Then f −1 : Spec Rred → Spec R is a homeomorphism. Exercise X.X: Prove Proposition 289. Exercise: Let R1 , . . . , Rn be finitely many rings. Show that Spec(R1 × . . . × Rn ) is ⨿n canonically homeomorphic to the topological space i=1 Spec Ri . Exercise: Let R be a Boolean ring. Earlier we defined a topology on the set “M (R)” of all maximal ideals of R. But, as we know, a Boolean ring all prime ideals are maximal, so as sets M (R) = Spec R. Show that moreover the topology we defined on M (R) is the Zariski topology on Spec R. 13.2. Properties of the spectrum: quasi-compactness. More than sixty years ago now, N. Bourbaki introduced the term quasi-compact for a topological space X for which any open covering has a finite subcovering. The point of this terminology is to reserve compact for a space which is both quasi-compact and Hausdorff, and thus emphasize that most of the nice properties of compact spaces in classical topology do rely on the Hausdorff axiom. Nowhere is this terminology more appropriate than in the class of spectral spaces, which as we have seen above, are only Hausdorff in the comparatively trivial case of a zero-dimensional ring. On the other hand: Proposition 290. For any commutative ring R, Spec R is quasi-compact. 178 PETE L. CLARK Proof. Let {Ui } be any open covering of Spec R. For each p ∈ Spec R, there exists an element U of the cover containing p, and thus a principal open set X (f ) containing p and contained in U . Therefore there is a refinement of the cover consisting of principal open subsets, and if this refinement has a finite cover, then the original cover certainly does as well. Thus it suffices to assume that the Ui ’s ∪ are basic open sets.45 So now suppose that Spec R = i X (fi ). Then we have ∪ ∪ ∩ Spec R = X (fi ) = (Spec R \ V (fi )) = Spec R \ V (fi ), i i i ∩ so that ∅ = i V (fi ) = V (⟨fi ⟩). Therefore the ideal I = ⟨fi ⟩ contains 1, and this means∩ that there is some finite subset f1 , . . . , fn of I such that ⟨f1 , . . . , fn ⟩ = R. ∪n n Thus i=1 V (fi ) = ∅, or equivalently, Spec R = i=1 X (fi ). 13.3. Properties of the spectrum: separation and specialization. For the reader’s convenience we briefly recall the “lower” separation axioms: A topological space X is Kolmogorov – or T0 – if for any distinct points x, y ∈ X , the system of neighborhoods Nx and Ny do not coincide. In plainer language, either there exists an open set U containing x and not containing y , or conversely. A topological space X is separated – or T1 – if for any distinct points x, y ∈ X , there exists both an open set U containing x and not y and an open set V containing y and not x. A space is separated iff all singleton sets {x} are closed iff for all ∩ x ∈ X , U ∈Nx U = {x}. A topological space X is Hausdorff – or T2 – if for any distinct points x, y ∈ X , there exist open neighborhoods U of x and V of y with U ∩ V = ∅. A space is Hausdorff iff for all x ∈ X , the intersection of all closed neighborhoods of x is {x}. Easily Hausdorff implies separated implies Kolmogorov. In a general topology course one learns that neither of the converse implications holds in general. On the other hand most of the spaces one encounters in analysis and geometry are Hausdorff, and certainly are if they are Kolmogorov (see Exercise X.X for a nonKolmogorov space arising naturally in analysis). We are about to see that yet a third state of affairs transpires when we restrict attention to spectra of rings. Let X be a topological space. We define a relation → on X by decreeing that for x, y ∈ X , x → y iff y lies in the closure of the singleton set {x}. This relation is called specialization, and we read x → y as “x specializes to y ”. The reader who is familiar with topology but has not seen the specialization relation before will find an explanation in part f) of the following exercise. Exercise X.X: a) Show that x → y iff Nx ⊂ Ny . 45This is just the familiar, and easy, fact that it suffices to verify quasi-compactness on any base for the topology. It is also true, but deeper, that one can verify quasi-compactness on any subbase: Alexander’s Subbase Theorem. COMMUTATIVE ALGEBRA 179 b) Show that specialization satisfies the following properties: (i) reflexivity: x → x; (ii) transitivity x → y, y → z =⇒ x → z . A relation R with these properties is called a quasi-ordering. Note that a partial ordering is a quasi-ordering with the additional axiom of anti-symmetry: xRy, yRx =⇒ x = y . c) Show that the specialization relation on X is a partial ordering iff X is Kolmogorov. d) Show that a point y is closed46 iff y → x =⇒ x = y . e) A point x for which x → y holds for all y ∈ X is called generic. Give an example of a topological space in which every point is generic. f) Show that X is separated iff x → y =⇒ x = y . Exercise X.X: Let X be any set endowed with a quasi-ordering R. Define a new relation x ≡ y if x R y and y R x. a) Show that ≡ is an equivalence relation on X . b) Write X ′ for the set of ≡ equivalence classes, and let q : X → X ′ be the natural map – i.e., x → {y ∈ X | y ≡ x}. Show that the relation R descends to a relation ≤ on X ′ : i.e., for s1 , s2 ∈ X ′ , then by choosing x1 ∈ s1 , x2 ∈ s2 and putting s1 ≤ s2 ⇐⇒ x1 R x2 , the relation ≤ is well-defined independent of the choices of x1 and x2 . Show that moreover ≤ is a partial ordering on X ′ . c) Let X be a topological space and R be the specialization relation. Endowing X ′ with the quotient topology via q , show that the induced relation ≤ on X ′ is the specialization relation on X ′ , and accordingly by the previous exercise X ′ is a Kolmogorov space. If it pleases you, show that q : X → X ′ is universal for maps from X into a Kolmogorov space Y , hence X ′ (or rather, q : X → X ′ ) can be regarded as the Kolmogorov quotient of X . Exercise: Let (X, µ) be a measure space, and let L1 be the space of all measurable ∫ ∫ functions f : X → R with X |f |dµ < ∞. For f ∈ L1 , define ||f || := X |f |dµ, and for ϵ > 0, put B (f, ϵ) = {g ∈ L1 | ||g − f || < ϵ}. Show that the B (f, ϵ)’s form a base for a topology on L1 , but that this topology is, in general, not Kolmogorov. Show that the Kolmogorov quotient is precisely the usual Lebesgue space L1 , whose elements are not functions but classes of functions modulo µ a.e. equivalence. Proposition 291. For any ring R, the spectrum Spec R is a Kolmogorov space. Indeed, for prime ideals p, q of R, we have p → q iff q ⊃ p, i.e., the specialization relation is precisely the opposite relation to the containment of prime ideals. Proof. For prime ideals p and q we have p → q ⇐⇒ q ∈ {p} = {f ∈ Spec R | f ⊃ p} ⇐⇒ q ⊃ p. Thus the specialization relation is just reverse containment of ideals, which certainly satisfies antisymmetry: q ⊂ p, p ⊂ q =⇒ p = q. Now apply Exercise X.Xc). 46Strictly speaking we mean that {y } is closed, but this abuse of terminology is very common and convenient. 180 PETE L. CLARK Lemma 292. For a ring R, TFAE: (i) R/ nil R is absolutely flat, i.e., every R/ nil R-module is flat. (ii) R has dimension zero: every prime ideal of R is maximal. Proof. We use the characterization of absolutely flat rings given in Proposition 180: it is necessary and sufficient that the localization at every maximal ideal be a field. Since the Krull dimension of Rm is the height of R, evidently any absolutely flat ring has Krull dimension 0. Since Spec(R/ nil R) = Spec R, dim R/ nil R = dim R, thus (i) =⇒ (ii). Conversely, localizations of reduced rings are reduced, and a reduced local ring of dimension zero is a field, so (ii) =⇒ (i). Theorem 293. For a commutative ring R, TFAE: (i) R/ nil R is absolutely flat, i.e., every R/ nil R-module is flat. (ii) R has Krull dimension zero. (iii) Spec R is a separated space. (iv) Spec R is a Hausdorff space. (v) Spec R is a Boolean space. Proof. The equivalence of (i) and (ii) is Lemma 292. (ii) ⇐⇒ (iii): a space is separated iff all of its singleton sets are closed. But if p is a prime ideal, then V (p) consists of all prime ideals containing p, so V (p) = {p} iff p is maximal. Certainly (v) =⇒ (iv) =⇒ (iii). (i) =⇒ (v): Since Spec R = Spec(R/ nil R), we may well assume that R itself is absolutely flat. Let p and q be distinct prime ideals; since both are maximal, there exists an element f ∈ p \ q. By Proposition 98, there exists an idempotent e with (e) = (f ), and therefore e ∈ p \ q. Then D(1 − e), D(e) is a separation of Spec R. More precisely, D(e) ∩ D(1 − e) = D(e(1 − e)) = D(e − e2 ) = D(0) = ∅, whereas for any prime ideal p, since 0 = e(1 − e) ∈ p, we must have e ∈ p or 1 − e ∈ pp. Moreover by construction we have p ∈ D(1 − e), q ∈ D(e). This shows that Spec R is Hausdorff, but in fact it shows more: given any two distinct points ⨿ P ̸= Q of the space X , we found a separation X = U V with P ∈ U, Q ∈ V , so that X is zero-dimensional. According to Proposition 290, every ring has quasicompact spectrum, so Spec R is Hausdorff, zero-dimensional and quasi-compact, i.e., Boolean. Exercise X.X: a) Let R be an arbitrary product of fields. Show that Spec R is a Boolean space. b) Let {Ri }i∈I be a family of rings, each of which has Krull dimension 0, and put ∏ R = i Ri . Must Spec R be Boolean? 13.4. Irreducible spaces. A topological space is irreducible if it is nonempty and if it cannot be expressed as the union of two proper closed subsets. Exercise X.X: Show that for a Hausdorff topological space X , TFAE: (i) X is irreducible. (ii) #X = 1. Proposition 294. For a topological space X , TFAE: (i) X is irreducible. COMMUTATIVE ALGEBRA 181 (ii) Every finite intersection of nonempty open subsets (including the empty intersection!) is nonempty. (iii) Every nonempty open subset of X is dense. (iv) Every open subset of X is connected. Exercise: Prove Proposition 294. Proposition 295. Let X be a nonempty topological space. a) If X is irreducible, every nonempty open subset of X is irreducible. b) If a subset Y of X is irreducible, so is its closure Y . c) If {Ui } is an open covering of X such that Ui ∩ Uj ̸= ∅ for all i, j and each Ui is irreducible, then X is irreducible. d) If f : X → Y is continuous and X is irreducible, then f (X ) is irreducible in Y . Proof. a) Let U be a nonempty open subset of X . By Proposition 294, it suffices to show that any nonempty open subset V of U is dense. But V is also a nonempty open subset of the irreducible space X . b) Suppose Y = A ∪ B where A and B are each proper closed subsets of Y ; since Y is itself closed, A and B are closed in X , and then Y = (Y ∩ A) ∪ (Y ∩ B ). If Y ∩ A = Y then Y ⊂ A and hence Y ⊂ A = A, contradiction. So A is proper in Y and similarly so is B , thus Y is not irreducible. c) Let V be a nonempty open subset of X . Since the Ui ’s are a covering of X , there exists at least one i such that V ∩ Ui ̸= ∅, and thus by irreducibility V ∩ Ui is a dense open subset of Ui . Therefore, for any index j , V ∩ Ui intersects the nonempty open subset Uj ∩ Ui , so in particular V intersects every element Uj of the covering. Thus for all sets Ui in an open covering, V ∩ Ui is dense in Ui , so V is dense in X . d) If f (X ) is not irreducible, there exist closed subsets A and B of Y such that A ∩ f (X ) and B ∩ f (X ) are both proper subsets of f (X ) and f (X ) ⊂ A ∪ B . Then f −1 (A) and f −1 (B ) are proper closed subsets of X whose union is all of X . Let x be a point of a topological space, and consider the set of all irreducible subspaces of X containing x. (Since {x} itself is irreducible, this set is nonempty.) The union of a chain of irreducible subspaces being irreducible, Zorn’s Lemma says that there exists at least one maximal irreducible subset containing x. A maximal irreducible subset (which, by the above, is necessarily closed) is called an irreducible component of X . Since irreducible subsets are connected, each irreducible component lies in a unique connected component, and each connected component is the union of its irreducible components. However, unlike connected components, it is possible for a given point to lie in more than one irreducible component. We will see examples shortly. In the case of a Zariski topology Spec R, there is an important algebraic interpretation of the irreducible components. Namely, the irreducible components Y of Spec R correspond to V (p) where p ranges through the minimal primes. Indeed, we first claim that a closed subset V (I ) is irreducible iff rad I = p is prime. First, if rad I = p is prime, then by Proposition 288 V (I ) = V (p), so that it suffices to show that V (p) is irreducible. If not, there exist ideals I and J such that V (I ) and V (J ) are both proper subsets of V (p) and V (p) = V (I ) ∪ V (J ) = V (IJ ). But then p = rad(IJ ) ⊃ IJ and since p is prime this implies either p ⊃ I or p ⊃ J . 182 PETE L. CLARK WLOG, suppose p ⊃ I ; then V (p) ⊂ V (I ), so that V (I ) is not proper in V (p), contradiction. Next, suppose that V (I ) is irreducible, and suppose that ab ∈ rad(I ). If neither a nor b is in rad(I ), then V ((a)), V ((b)) do not contain V (I ), but V (a) ∪ V (b) = V (ab) ⊃ V (I ). Therefore V (a) ∩ V (I ) = V (aI ) and V (b) ∩ V (I ) = V (bI ) are two proper closed subsets of V (I ) whose union is V (I ), thus V (I ) is reducible. It follows that the irreducible components – i.e., the maximal irreducible subsets – are precisely the sets of the form V (p) as p ranges over the distinct minimal prime ideals. Note that we can now deduce that minimal prime ideals exist in any ring as a special case of the existence of irreducible components in any topological space, an example of the use of topological methods to prove purely algebraic results.47 Proposition 296. For any commutative ring, the map p → V (p) gives a bijection from Spec R to the set of irreducible closed subsets of Spec R. Exercise X.X: Prove Proposition 296. Exercise: Explain why Proposition 296 is, in some sense, a Nullstellensatz for an arbitrary commutative ring. 13.5. Noetherian spaces. We wish to introduce a property of topological spaces which, from the standpoint of conventional geometry, looks completely bizarre: Proposition 297. For a topological space X , TFAE: (i) Every ascending chain of open subsets is eventually constant. (ibis) Every descending chain of closed subsets is eventually constant. (ii) Every nonempty family of open subsets has a maximal element. (iibis) Every nonempty family of closed subsets has a minimal element. (iii) Every open subset is quasi-compact. (iv) Every subset is quasi-compact. A space satisfying any (and hence all) of these conditions is called Noetherian. Proof. The equivalence of (i) and (ibis), and of (ii) and (iibis) is immediate from taking complements. The equivalence of (i) and (ii) is a general property of partially ordered sets discussed in section X.X above. (i) ⇐⇒ (iii): Assume (i), let U be any open set in X and let {Vj } be an open covering of U . We assume for a contradiction that there is no finite subcovering. Choose any j1 and put U1 := Vj1 . Since U1 ̸= U , there exists j2 such that U1 does not contain Vj2 , and put U2 = U1 ∪ Vj2 . Again our assumpion implies that U2 U , and continuing in this fashion we will construct an infinite properly ascending chain of open subsets of X , contradiction. Conversely, assume (iii) and let {Ui }∞ be an i=1 ∪ infinite properly ascending chain of subsets. Then U = i Ui is not quasi-compact. Obviously (iv) =⇒ (iii), so finally we will show that (iii) =⇒ (iv). Suppose that Y ⊂ X is not quasi-compact, and let {Vi }i∈I be a covering of Y by relatively open subsets without a finite subcover. We may write each Vi as Ui ∩ Y with Ui open 47In this case, the same Zorn’s Lemma argument establishes the existence of minimal primes, so the topology is not making anything essentially easier for us...yet. COMMUTATIVE ALGEBRA 183 ∪ in Y . Put U = i Ui . Then, since U is quasi-compact, there exists a ∪ finite subset ∪ ∪ J ⊂ I such that U = j ∈J Uj , and then Y = U ∩ Y = j ∈J Uj ∩ Y = j ∈J Vj . Corollary 298. A Noetherian Hausdorff space is finite. Proof. In a Hausdorff space every quasi-compact subset is closed. Therefore, using the equivalence (i) ⇐⇒ (iv) in Proposition 297, in a Noetherian Hausdorff space every subset is closed, so such a space is discrete. But it is also quasi-compact, so it is finite. Proposition 299. For a ring R, TFAE: (i) R satisfies the ascending chain condition on radical ideals. (ii) Spec R is a Noetherian space. In particular if R, or even Rred = R/ nil(R), is a Noetherian ring, Spec R is a Noetherian space. Proof. Since I → V (I ) gives a bijection between radical ideals and Zariski closed subsets, (ACC) on radical ideals is equivalent to (DCC) on closed subsets. Evidently these conditions occur if R is itself Noetherian, or, since Spec R is canonically homeomorphic to Spec Rred , if Rred is Noetherian. Proposition 300. Let X be a Noetherian topological space. ∪n a) There are finitely many closed irreducible subsets {Ai }n such that X = i=1 Ai . i=1 b) Starting with any finite family {Ai }n as in part a) and eliminating all reduni=1 dant sets – i.e., all Ai such that Ai ⊂ Aj for some j ̸= i – we arrive at the set of irreducible components of X . In particular, the irreducible components of a Noetherian space are finite in number. Proof. a) Let X be a Noetherian topological space. We first claim that X can be expressed as a finite union of irreducible closed subsets. Indeed, consider the collection of closed subsets of X which cannot be expressed as a finite union of irreducible closed subsets. If this collection is nonempty, then by Proposition 297 there exists a minimal element Y . Certainly Y is not itself irreducible, so is the union of two strictly smaller closed subsets Z1 and Z2 . But Z1 and Z2 , being strictly smaller than Y , must therefore be expressible as finite unions of irreducible closed subsets and therefore so also can Y be so expressed, contradiction. b) So write X = A1 ∪ . . . ∪ An where each Ai is closed and irreducible. If for some i ̸= j we have Ai ⊂ Aj , then we call Ai redundant and remove it from our list. After a finite number of such removals, we may assume that the above finite covering of X by closed irreducibles is irredundant in the sense that there are no containment relations between distinct ∪n Ai ’s. Now let Z be any irreducible closed subset. Since Z = i=1 (Z ∩ Ai ) and Z is irreducible, we must have Z = Z ∩ Ai for some i, i.e., Z ⊂ Ai . It follows that the “irredundant” Ai ’s are precisely the maximal irreducible closed subsets, i.e., the irreducible components. We deduce the following important result, which is not so straightforward to prove using purely algebraic methods: Corollary 301. Let I be a proper ideal in a Noetherian ring R. The set of prime ideals p which are minimal over I (i.e., minimal among all prime ideals containing I ) is finite and nonempty. 184 PETE L. CLARK Exercise: Prove Corollary 301. 13.6. The map on spectra induced by a homomorphism. 13.7. Hochster’s Theorem. A topological space X is sober if for every irreducible closed subspace Y of X , there exists a unique point y ∈ Y such that Y = {y }. Equivalently, a sober space is one for which every irreducible closed subset has a unique generic point. Exercise X.X: a) Show that any Hausdorff space is sober. b) Show that a sobser space is Kolmogorov. c) Show that the cofinite topology on an infinite set is separated but not sober. Exercise (Sobrification): For any topological space X , let X t be the set of irreducible closed subsets of X . There is a natural map t : X → X t via x → {x}. Give X t the final topology with respect to t, i.e., the finest topology that makes t continuous. (Explicitly, a subset V of X t is open iff its preimage in X is open.) a) Show the map t induces a bijection from the open subsets of X to the open subsets of X t . b) Show that X t is a sober space. c) Show that t is universal for continuous maps from X to a sober topological space: i.e., for every sober space Y and continuous f : X → Y , there exists a unique continuous F : X t → Y such that f = F ◦ t. Thus X t is (unfortunately!) called the sobrification of X . A topological space X is spectral if: (SS1) X is quasi-compact, (SS2) X is sober, and (SS3) The family of quasi-compact open subsets of X is closed under finite intersections and is a base for the topology. Remark: A Bourbakiste would insist that (SS3) =⇒ (SS1) by taking the empty intersection. But we will not do so. Exercise: Show that a finite space is spectral iff it is T0 . The following result gives an arguably cleaner characterization of spectral spaces. Proposition 302. For a topological space X , TFAE: (i) X is homeomorphic to an inverse limit of finite T0 spaces. (ii) X is spectral. Exercise X.X: Prove Proposition 302.48 Proposition 303. For any ring R, Spec R is spectral. 48I presume this is a relatively easy exercise working directly from the definitions, but I have not attempted it myself. Please contact me if you have tried to prove this result but failed. COMMUTATIVE ALGEBRA 185 Exercise: Prove Proposition 303. (Hint: you will find the needed results in the previous subsections. Especially, use Proposition 296 to prove the sobriety of Spec R.) For any ring R we endow the set MaxSpec(R) of maximal ideals of R with the topology it inherits as a subset of Spec(R). When necessary, we describe MaxSpec R as the “maximal spectrum” of R. Proposition 304. For any ring R, MaxSpec R is separated and quasi-compact. Theorem 305. (Hochster’s Thesis [Ho69]) a) A spectral topological space is homeomorphic to the prime spectrum of some ring. b) A separated quasi-compact space is homeomorphic to the maximal spectrum of some ring. We do not aspire to give a proof of Theorem 305 at this time. (If you know of a nice, short proof, please tell me!) Exercise: Show that every compact (this includes Hausdorff!) space X is homeomorphic to MaxSpec(C (X )), where C (X ) is the ring of continuous R-valued functions on X . Exercise: a) Show that the specialization relation gives an equivalence of categories between the category of T0 finite spaces and the category of finite partially ordered sets. b)* Formulate a generalization of part a) in which T0 finite spaces are replaced by T0 Alexandroff spaces. (A topological space is Alexandroff if an arbitrary intersection of closed subsets is closed.) Exercise: Let n ∈ Z+ . a) Use Hochster’s Thesis and the previous exercise to show that there exists a ring R with exactly n prime ideals p1 , . . . , pn such that p1 ⊂ p2 ⊂ . . . ⊂ pn . b) For n = 1, 2, exhibit Noetherian rings with these properties. For n ≥ 3, show that there is no such Noetherian ring. 13.8. The support of a module. Let M be a module over a ring R. For m ∈ M , we put ann m = {x ∈ R | xm = 0}. Also we define ∩ ann M = {x ∈ R | xM = 0} so that ann M = m∈M ann m. (We have seen this concept before: we defined a module to be faithful if ann M = 0.) Exercise X.X: ∩ {ωi }i∈I be a set of generators for M as an R-module. Show Let that ann M = i∈I ann ωi . For a module M over a ring R, we define its support supp M = {p ∈ Spec R | Mp ̸= 0}. 186 PETE L. CLARK This is an extremely important geometric notion which alas, gets very short shrift in these notes. But we will establish the following basic result. Proposition 306. For a finitely generated R-module M , supp M = V (ann M ). In particular, supp M is a Zariski closed set. Proof. Write M = ⟨ω1 , . . . , ωn ⟩R . For p ∈ Spec R, we have p ∈ supp M iff Mp ̸= 0 iff there exists i such that the image of ωi in Mp is not zero iff there exists i such that ann(ωi ) ⊂ p iff n ∩ ann M = ann(ωi ) ⊂ p. i=1 13.9. Rank functions revisited. Theorem 307. Let M be a finitely generated module over a ring R. a) For each n ∈ N, the set Ur = {p ∈ Spec R | Mp can be generated over Rp by at most r elements} is open in Spec R. b) If M is finitely presented (e.g. if R is Noetherian), then the set UF = {p ∈ Spec R | Mp is a free Rp -module} is open in Spec R. Proof. (Matsumura) Suppose Mp = ⟨ω1 , . . . , ωr ⟩Rp . Each ωi is of the form mii with s × mi ∈ M and si ∈ R \ p. But since si ∈ Rp for all i, we also have ⟨m1 , . . . , mr ⟩Rp = Mp . Thus it is no loss of generality to assume that each ωi is the image in Mp of an element of M . Let φ : Rr → M be the R-linear map given by (a1 , . . . , ar ) → ∑ i ai ωi , and put C = coker φ, whence an exact sequence Rr → M → C → 0. Localizing this at a prime q of R gives an exact sequence r Rq → Mq → Cq → 0. When q = p we of course have Cq = 0. Moreover, C is a quotient of M hence a finitely generated R-module, so by Proposition 306 its support supp C is a Zariskiclosed set. It follows that there exists an open neighborhood V of p such that Cq = 0 for all q ∈ V . b) Suppose that Mp is a free Rp -module with basis ω1 , . . . , ωr . As above it is no loss of generality to assume that each ωi is the image in Mp of an element of M . Moreover, as we have also just seen, there exists a basic open neighborhood U (f ) such that for all q ∈ U (f ), the images of ω1 , . . . , ωr in Mq generate Mq as an Rq module. Replacing R by Rf and M by Mf we may assume that this occurs for all q ∈ Spec R. Thus M/⟨ω1 , . . . , ωr ⟩R is everywhere locally zero, so it is locally zero: M = ⟨ω1 , . . . , ωr ⟩. Defining an R-linear map φ : Rr → M as above and setting K = Ker φ, we have the exact sequence 0 → K → Rr → M → 0. COMMUTATIVE ALGEBRA 187 Since M is finitely presented, according to Proposition 14 K is a finitely generated R-module. Moreover we have Kp = 0 hence as above Kq = 0 for all q on some open neighborhood V of p. By construction, for each q ∈ V , the images of ω1 , . . . , ωr in Mq give an Rq -basis for Mq . Let M be a finitely generated, locally free module over a ring R. Earlier we defined the rank function r : Spec R → N as the rank of the free, finitely generated Rp module Mp . Applying Theorem 307a) to the locally free module M says precisely that the rank function is lower-semicontinuous : it can jump up upon specialization, but not jump down. We now ask the reader to look back at Theorem 183 and see that for a finitely generated module M over a general ring R, M is projective iff it is locally free and finitely presented. When R is Noetherian, being finitely presented is equivalent to being finitely generated, so being projective is the same as being locally free. However, in the general case we have had little to say about the distinction between finitely presented and finitely generated modules. Is there some way to rephrase the subtly stronger property of finite presentation, perhaps a more geometric way? Indeed there is: Theorem 308. Let M be a finitely generated locally free R-module. TFAE: (i) The rank function rM : Spec R → N is locally constant. (ii) M is a projective module. Proof. (i) =⇒ (ii): By Theorem 183, it is enough to show that for all m ∈ MaxSpec R, there exists f ∈ R \ m such that Mf is a free module. Let n = r(m), and let x1 , . . . , xn be an Rm -basis for Mm . Choose X1 , . . . , Xn ∈ M such that for × all i, the image of Xi in Mm is of the form ui xi for uu ∈ Rm . Let u : Rn → M be the map sending the ith standard basis element ei to Xi . Since M is finitely n generated, by Proposition 181 there exists f ∈ R \ m such that uf : Rf → Mf is surjective. It follows that for all g ∈ R \ m, uf g is surjective. Moreover, by hypothesis there is some such g such that r(p) = n for all p ∈ X (g ). Replacing f n by f g we may assume that r(p) = n for all p ∈ X (f ). For all such p, up : Rp → Mp is therefore a surjective endomorphism from a rank n free module to itself. Since finitely generated modules are Hopfian, up is an isomorphism. By the local nature of isomorphisms (Proposition 175) we conclude uf is an isomorphism, so Mf is free. (ii) =⇒ (i): By Theorem 183, M is Z-locally free: there exists a finite Z-family {fi }∏ I such that for all i ∈ I , Mfi is finitely generated and free. Thus the modi∈ n ule i=1 Mfi is finitely generated and projective over the faithfully flat R-algebra ∏n i=1 Rfi , so by faithfully flat descent (Theorem 109) M itself is projective. Corollary 309. Let R be a ring with Spec R irreducible (e.g. a domain). For a finitely generated R-module M , TFAE: (i) R is projective. (ii) R is Z-locally free. (iii) R is locally free. (iv) R is flat. Exercise: Prove Corollary 309. 188 PETE L. CLARK 14. Integrality in Ring Extensions 14.1. First properties of integral extensions. If S is a ring extension of R – i.e., R ⊂ S – we will say that an element α of S is integral over R if there exist a0 , . . . , an−1 ∈ R such that αn + an−1 αn−1 + . . . + a1 α + a0 = 0. Note that every element α ∈ R satisfies the monic polynomial t − α = 0, so is integral over R. Theorem 310. Let R ⊂ T be an inclusion of rings, and α ∈ T . TFAE: (i) α is integral over R. (ii) R[α] is finitely generated as an R-module. (iii) There exists an intermediate ring R ⊂ S ⊂ T such that α ∈ S and S is finitely generated as an R-module. (iv) There exists a faithful R[α]-submodule M of T which is finitely generated as an R-module. Proof. (i) =⇒ (ii): If α is integral over R, there exist a0 , . . . , an−1 ∈ R such that αn + an−1 αn−1 + . . . + a1 α + a0 = 0, or equivalently αn = −an−1 αn−1 − . . . − a1 α − a0 . This relation allows us to rewrite any element of R[α] as a polynomial of degree at most n − 1, so that 1, α, . . . , αn−1 generates R[α] as an R-module. (ii) =⇒ (iii): Take T = R[α]. (iii) =⇒ (iv): Take M = S . (iv) =⇒ (i): Let m1 , . . . , mn be a finite set of generators for M over R, and express each of the elements mi α in terms of these generators: αmi = n ∑ rij mj , rij ∈ R. j =1 Let A be the n × n matrix αIn − (rij ); then recall from linear algebra that AA∗ = det(A) · In , where A∗ is the “adjugate” matrix (of cofactors). If m = (m1 , . . . , mn ) (the row vector), then the above equation implies 0 = mA = mAA∗ = m det(A) · In . The latter matrix equation amounts to mi det(A) = 0 for all i. Thus • det(A) = •0 on M , and by faithfulness this means det(A) = 0. Since so that α is a root of the monic polynomial det(T · In − (aij )). Exercise: Let S be a finitely generated R-algebra. Show that TFAE: (i) S/R is integral. (ii) S is finite over R (as an R-module!). In particular, if S/R is an extension and α1 , . . . , αn are all integral over R, then R[α1 , . . . , αn ] is a finitely generated R-module. COMMUTATIVE ALGEBRA 189 Proposition 311. (Integrality is preserved under quotients and localizations) Let S/R be an integral ring extension. a) Let J be an ideal of S . Then S/J is an integral extension of R/(J ∩ R). b) Let T be a multiplicatively closed subset of nonzero elements of R. Then ST is an integral extension of RT . Proof. a) First note that the kernel of the composite map R → S → S/J is J ∩ R, so that R/(J ∩ R) → S/J is indeed a ring extension. Any element of S/J is of the form x + J for x ∈ S , and if P (t)tn + an−1 tn−1 + . . . + a1 t + a0 = 0 ∈ R[t] is a polynomial satisfied by x, then reducing coefficientwise gives a monic polynomial P (t) ∈ R/(J ∩ R) satisfied by x. b) Let J = {s ∈ S |∃t ∈ T | ts = 0}, an ideal of S . Then ST is a quotient ring of S/J and RT is a quotient ring of R/(J ∩ R), by part a) we are reduced to the case in which R → RT and S → ST . Let x , x ∈ S, y ∈ T be an element of ST . Let y P (t) = tn + an−1 tn−1 + . . . + a0 ∈ R[t] be a monic polynomial satisfied by S . Then ( )n ( )n−1 x an−1 x a0 + + . . . + n = 0, y y y y showing that x y is integral over RT . Lemma 312. Let R ⊂ S ⊂ T be an inclusion of rings. If α ∈ T is integral over R, then it is also integral over S . Proof. If α is integral over R, there exists a monic polynomial P ∈ R[t] such that P (α) = 0. But P is also a monic polynomial in S [t] such that P (α) = 0, so α is also integral over S . Lemma 313. Let R ⊂ S ⊂ T be rings. If S is finitely generated as an R-module and T is finitely generated as an S -module then T is finitely generated as an Rmodule. Proof. If α1 , . . . , αr generates S as an R-module and β1 , . . . , βs generates T as an S -module, then {αi βj }{ 1 ≤ i ≤ r, 1 ≤ j ≤ s} generates T as an R-module: for α ∈ T , we have ∑ ∑∑ α= bj βj = (aij αi )βj , j i j with bj ∈ S and aij ∈ R. Corollary 314. (Transitivity of integrality) If R ⊂ S ⊂ T are ring extensions such that S/R and T /S are both integral, then T /R is integral. Proof. For α ∈ S , let αn + bn−1 αn−1 + . . . + b1 α + b0 = 0 be an integral dependence relation, with bi ∈ S . Thus R[b1 , . . . , bn−1 , α] is finitely generated over R[b1 , . . . , bn−1 ]. Since S/R is integral, R[b1 , . . . , bn−1 ] is finite over R. By Lemma 313, R[b1 , . . . , bn−1 , α] is a subring of T containing α and finitely generated over R, so by Theorem 310, α is integral over R. Corollary 315. If S/R is a ring extension, then the set IS (R) of elements of S which are integral over R is a subring of S , the integral closure of R in S . Thus R ⊂ IS (R) ⊂ S . 190 PETE L. CLARK Proof. If α ∈ S is integral over R, R[α1 ] is a finitely generated R-module. If α2 is integral over R it is also integral over R[α1 ], so that R[α1 ][α2 ] is finitely generated as an R[α1 ]-module. By Lemmma 313, this implies that R[α1 , α2 ] is a finitely generated R-module containing α1 ± α2 and α1 · α2 . By Theorem 310, this implies that α1 ± α2 and α1 α2 are integral over R. If R ⊂ S such that IS (R) = R, we say R is integrally closed in S . Proposition 316. Let S be a ring. The operator R → IS (R) on subrings of R is a closure operator in the abstract sense, namely it satisfies: (CL1) R ⊂ IS (R), (CL2) R1 ⊂ R2 =⇒ IS (R1 ) ⊂ IS (R2 ). (CL3) IS (IS (R)) = IS (R). Proof. (CL1) is the (trivial) Remark 1.1. (CL2) is obvious: evidently if R1 ⊂ R2 , then every element of S which satisfies a monic polynomial with R1 -coefficients also satisfies a monic polynomial with R2 -coefficients. Finally, suppose that α ∈ S is such that αn + an−1 αn−1 + . . . + a1 α + a0 = 0 for ai ∈ IS (R). Then each ai is integral over R, so R[a1 , . . . , an ] is finitely generated as an R-module, and since R[a1 , . . . , an , α] is finitely generated as an R[a1 , . . . , an ]-module, applying Lemma 313 again, we deduce that α lies in the finitely generated R-module R[a1 , . . . , an , α] and hence by Theorem 310 is integral over R. 14.2. Integral closure of domains. Until further notice we restrict to the case in which R ⊂ S are integral domains. Proposition 317. Let R ⊂ S be an integral extension of integral domains. a) R is a field iff S is a field. b) An extension of fields is integral iff it is algebraic. Proof. a) Suppose first that R is a field, and let 0 ̸= α ∈ S . Since α is integral over R, R[α] is finitely generated as an R-module, and it is well-known in field theory that this implies R[α] = R(α). Indeed, taking the polynomial of least degree satisfied by α, say α(αn−1 + an−1 αn−2 + . . . + a1 ) = −a0 , then 0 ̸= a0 ∈ R is invertible, so −1 n−1 (α + an−1 αn−2 + . . . + a1 ) = α−1 , a0 and S is a field. Conversely, if S is a field and a ∈ R, then R[a−1 ] is finitedimensional over R, i.e., there exist ai ∈ R such that a−n = an−1 a−n+1 + . . . + a1 a−1 + a0 . Multiplying through by an−1 gives a−1 = an−1 + an−2 a + . . . + a1 an−2 + a0 an−1 ∈ R, completing the proof of part a). Over a field every polynomial relation can be rescaled to give a monic polynomial relation, whence part b). Remark: A more sophisticated way of expressing Proposition 317 is that if S/R is an integral extension of domains, then dim R = 0 iff dim S = 0. Later we will see that in fact dim R = dim S under the same hypotheses. The integral closure of one field R in another field S is called the algebraic closure COMMUTATIVE ALGEBRA 191 of R in S . Exercise: a) Let S/R be an extension of fields. If S is algebraically closed, then so is IS (R). b) Deduce that if R = Q, S = C, then IS (R) is an algebraically closed, algebraic extension of Q, denoted Q and called the field of all algebraic numbers. Theorem 318. Let S/R be an extension of integral domains, and let T ⊂ R be a multiplicatively closed subset. Then IT −1 S (T −1 R) = T −1 IS (R). In other words, localization commutes with integral closure. Proof. Let K be the fraction field of R and L the fraction field of S . Then T −1 IS (R) is precisely the subring of L generated by T −1 and the elements of S which are integral over R. Since both of these kinds of elements of T −1S are integral over T −1 R and integral elements form a subring, we must have T −1 IS (R) ⊂ IT −1 S (T −1 R). Conversely, let s ∈ T −1 S be an element which is integral over T −1 R, so that there t exist a0 , . . . , an−1 ∈ T −1 R such that s s ( )n + an−1 ( )n−1 + . . . + a0 = 0. t t Clearing denominators, we get sn + (tan−1 )sn−1 + . . . + tn = 0, which shows that s is integral over R, so s t ∈ T −1 IS (R). Proposition 319. Suppose S/R is an extension with R a domain and S a field. Let K be the fraction field of R and M the fraction field of S . Then the fraction field of IS (R) is IM (K ). Proof. Let us write L for the quotient field of IS (R). First we show IM (K ) ⊂ L: let 0 ̸= x ∈ IS (K ), so there exist a0 , . . . , an−1 ∈ K such that xn + an−1 xn−1 + . . . + a1 x + a0 = 0. After clearing denominators and relabelling, we get a0 , . . . , an ∈ R such that an xn + an−1 xn−1 + . . . + a1 x + a0 = 0. Multiplying through by an−1 , we get n (an x)n + an−1 (an x)n−1 + . . . + a1 an−2 (an x) + an−1 a0 = 0, n n which shows that an x ∈ IS (R), so x ∈ IS (R) · (R \ 0)−1 hence is in the fraction field of IS (R). The reverse inclusion L ⊂ IM (K ) is quite similar, since an arbitrary nonzero element x of L is of the form α with α, β integral over R. But then certainly α and β β are both integral over K – i.e., algebraic over K , and then so also are β −1 and x = α . So again x satisfies a polynomial an tn + . . . with coefficients in R and then β an x is integral over R, hence an x and an are algebraic over K and thus x = an x is x algebraic over K so lies in IM (K ). Example: If R = Z, S = C, then IS (R) is called the ring of all algebraic integers, and often denoted Z. By Proposition 319, its fraction field is the field Q of all algebraic numbers. This turns out to be a very interesting ring, and it will crop up several times in the sequel as an example or counterexample. For instance: 192 PETE L. CLARK Exercise 1.1.3: Show that Z is not finitely generated as a Z-module.49 This shows that an integral extension need not be finite but only “locally finite” (here we are using the word “locally” in a rough intuitive sense. not in the sense of localization in commutative algebra). Corollary 320. Let R be a domain with fraction field K and M/K an arbitrary field extension. Put S = IM (R). Then S is integrally closed. Proof. By Proposition 319, the field of fractions L of S is the algebraic closure of K in M . If x ∈ L is integral over R, then since L ⊂ M it lies in IM (R) = S . Remark on terminology: It is unfortunate that the word “integral” in commutative algebra is used both to describe rings without zero divisors and also – quite distinctly – to describe an extension satisfying a certain kind of finiteness condition. The first use of the word “integral” is essentially redundant in the algebraic setting: if we just said “domain” instead of “integral domain”, the meaning would be the same. However, in geometric language one has the notion of an “integral scheme”; in the case of the affine scheme Spec R associated to a commutative ring R, the meaning of this is precisely that R be a domain, and because of this it is undesirable to banish the “integral” from “integral domain.” Perhaps it would be better instead to replace the term “integral extension.” We leave it as an exercise to the reader to suggest some alternate terminology: this is less silly than it sounds, because in order to do this one needs to grapple with the question, “What is an integral extension, really?” It is perhaps telling that there is, so far as I know, no notion of an “integral scheme extension” in algebraic geometry. Let R be an integral domain with fraction field K . We say that R is integrally closed if IK (R) = R, i.e., if any element of the fraction field satisfying a monic integral polynomial with R-coefficients already belongs to R. It follows immediately from Proposition 316 that in any case IK (R) is integrally closed. √ Exercise 1.1.4: Let R = Z[ −3] = Z[t]/(t2 + 3). Show that R is not integrally closed, and compute its integral closure. The geometric terminology for an integrally closed domain is normal. The process of replacing R by its integral closure IK (R) is often called normalization. 14.3. Spectral properties of integral extensions. Going down (GD): If we have I1 ⊃ I2 of R and J1 ∈ Spec S such that J1 ∩ R = I1 , there exists J2 ∈ Spec S such that J2 ⊂ J1 and J2 ∩ R = I2 . Lemma 321. Let R be a local ring with maximal ideal p and S/R an integral extension. Then the pushed forward ideal pS is proper. ∑ Proof. Suppose not: then there exist pi ∈ p, si ∈ S such that 1 = i si pi . Therefore any counterxample would take place already in the finite R-module 49In fact it is not even a Noetherian ring, so not even finitely generated as a Z-algebra. COMMUTATIVE ALGEBRA 193 R[s1 , . . . , sd ]. By induction on d, it is enough to consider the case of n = 1: S = R[s]. Consider as usual a relation (17) sn = an−1 sn−1 + . . . + a1 s + a0 , ai ∈ R of minimal possible degree n. If 1 ∈ pS then we have (18) 1 = p0 + p1 s + . . . + pk sk , pi ∈ p. In view of (17) we may assume k ≤ n − 1. Since 1 − p0 is not in the maximal ideal of the local ring R, it is therefore a unit; we may therefore divide (18) by 1 − p0 and get an equation of the form 1 = p′ s + . . . + p′ sq , p′ ∈ p. 1 q i This shows that s ∈ S × . Replacing a0 = a0 · 1 in (17) by a0 (p′ s + . . . + p′ sq ), we q 1 get an integral dependence relation which is a polynomial in s with no constant term. Since s is a unit, we may divide through by it and get an integral dependence relation of smaller degree, contradiction. Theorem 322. An integral ring extension S/R satisfies property (LO):50 every prime ideal p of R is of the form S ∩ P for a prime ideal P of S . Proof. For p a prime ideal of R, we denote – as usual – by Rp the localization of R at the multiplicatively closed subset R \ p. Then Rp is local with unique maximal ideal pRp , and if we can show that there exists a prime ideal Q of Sp lying over pRp , then the pullback P = Q ∩ S to S is a prime ideal of S lying over p. By Lemma 321, there exists a maximal ideal Q ⊃ pS and then Q ∩ R is a proper ideal containing the maximal ideal p and therefore equal to it. Corollary 323. (Going up theorem of Cohen-Seidenberg [CS46]) Let S/R be an integral extension and p ⊂ q be two prime ideals of R. Let P be a prime ideal of S lying over p (which necessarily exists by Theorem 322). Then there exists a prime ideal Q of S containing P and lying over q. Proof. Apply Theorem 322 with R = R/p S = S/P and p = q/p. Corollary 324. (Incomparability) Suppose S/R is integral and P ⊂ Q are two primes of S . Then P ∩ R ̸= Q ∩ R. Proof. By passage to S/P , we may assume that P = 0 and S is an integral domain, and our task is to show that any nonzero prime ideal P of S lies over a nonzero ideal of R. Indeed, let 0 ̸= x ∈ P , and let P (t) = tn + an−1 tn−1 + . . . + a0 ∈ R[t] be a monic polynomial satisfied by x; we may assume a0 ̸= 0 (otherwise divide by t). Then a0 ∈ xS ∩ R ⊂ P ∩ R. Corollary 325. Let S/R be an integral extension, P a prime ideal of S lying over p. Then P is maximal iff p is maximal. Proof. First proof: Consider the integral extension S/P /(R/p); the assertion to be proved is that S/P is a field iff R/p is a field. But this is precisely Proposition 317a). Second proof: If p is not maximal, it is properly contained in some maximal ideal q. By the Going Up Theorem, there exists a prime Q ⊃ P lying over q, so P is not maximal. Conversely, suppose that p is maximal but P is not, so there 50Or, lying over. 194 PETE L. CLARK exists Q P . Then Q ∩ R is a proper ideal containing the maximal ideal p, so Q ∩ R = p = P ∩ R, contradicting the Incomparability Theorem. Invoking Going Up and Incomparability to (re)prove the elementary Corollary 325 is overkill, but these more sophisticated tools also prove the following Corollary 326. Let S/R be an integral extension of rings. Then the Krull dimensions of R and S are equal. Proof. Recall that the Krull dimension of a ring is the supremum of the length of a finite chain of prime ideals. Suppose p0 p1 . . . pd are primes in R. Applying Theorem 322, we get a prime P0 of S lying over p0 , and then repeated application of the Going Up Theorem yields a chain of primes P0 P1 ... Pd , so that dim(S ) ≥ dim(R). Similarly, if we have a chain of prime ideals P0 . . . Pd of length d in S , then the Incomparability Theorem 324 implies that for all 0 ≤ i < d, Pi ∩ R Pi+1 . 14.4. Integrally closed domains. Let S/R be an extension of integral domains. Immediately from the definition of integrality, there is a very concrete way to show that an element x ∈ S is integral over R: we just have to exhibit a monic polynomial P ∈ R[t] such that P (x) = 0. What if we want to show that an algebraic element x is not integral over R? It would suffice to show that R[x] is not a finite R-module, but exactly how to do this is not clear. √ As an example, it is obvious that α = 2 is an algebraic integer, but unfortu√ nately it is not obvious that β = 22 is not an algebraic integer. (And of course we √ need to be careful, because e.g. γ = 1+2 5 is an algebraic integer, since it satisfies t2 + t − 1 = 0.) One thing to notice is that unlike α and γ , the minimal polynomial of β , t2 − 1 , does not have Z-coefficients. According to the next result, this is 2 enough to know that β is not integral over Z. Theorem 327. Let R be a domain with fraction field K , S/R an extension ring, and x ∈ S an integral element over R. a) Let P (t) ∈ K [t] be the minimal polynomial of x over K . Then P (t) ∈ IK (R)[t]. b) So if R is integrally closed, the minimal polynomial of x has R-coefficients. Proof. a) We may assume without loss of generality that S = R[x]; then, by integrality, S is a finite R-module. Let L be the fraction field of S , d = [L : K ], and let s1 , . . . , sd be a K -basis for L consisting of elements of S . Because the minimal polynomial P (t) is irreducible over K , it is also the characteristic polynomial of x· viewed as a K -linear automorphism of L. Evidently the matrix Mx of x· with respect to the basis s1 , . . . , sd has coefficients in S ∩ K , hence in IK (R). The coefficients of P (t) are polynomials in the entries of the matrix Mx , hence they also lie in IK (R). Part b) follows immediately. Exercise X.X.X: Let R be a domain with fraction field K . Let S/R be an extension such that for every x ∈ S which is integral over R, the minimal polynomial P (t) ∈ K [t] has R-coefficients. Show that R is integrally closed. COMMUTATIVE ALGEBRA 195 Theorem 328. (Local nature of integral closure) For an integral domain R, TFAE: (i) R is integrally closed. (ii) For all prime ideals p of R, Rp is integrally closed. (iii) For all maximal ideals m of R, Rm is integrally closed. Proof. Let K be the fraction field of R. Assume (i), and let p ∈ Spec R. By Theorem 318, the integral closure of Rp in K is Rp . Evidently (ii) =⇒ (iii). Assume (iii), and let x be an element of K which is integral over R. Then for every maximal ideal m of R, certainly x is integral over Rm , so by assumption x ∈ Rm ∩ ∩ and thus x ∈ m Rm . By Corollary 176 we have m Rm = R. Exercise X.X: Let R be an integrally closed domain with fraction field K , L/K an algebraic field extension, S the integral closure of R in L and G = Aut(L/K ). a) Show that for every σ ∈ G, σ (S ) = S . b) For P ∈ Spec S and σ ∈ G, show σ (P ) = {σ (x) | x ∈ P} is a prime ideal of S . c) Show that P ∩ R = σ (P ) ∩ R. In conclusion, for every p ∈ Spec R, there is a well-defined action of G on the (nonempty!) set of prime ideals P of S lying over p. Lemma 329. Let R be a domain with fraction field K of characteristic p > 0, let L/K be a purely inseparable algebraic extension of K (possibly of infinite degree), and let S be the integral closure of R in L. For any p ∈ Spec R, rad(pR) is the unique prime of S lying over p. Exercise: Prove Lemma 329. (Suggestions: recall that since L/K is purely insepa arable, for every x ∈ L, there exists a ∈ N such that xp ∈ K . First observe that rad(pR) contains every prime ideal of S which lies over p and then show that rad(pR) is itself a prime ideal.) Theorem 330. (Going Down Theorem of Cohen-Seidenberg [CS46]) Let R be an integrally closed domain with fraction field K , and let S be an integral extension of R. If p1 ⊂ p2 are prime ideals of R and P2 is a prime ideal of S lying over p2 , then there exists a prime ideal P1 of S which is contained in P2 and lies over p1 . Proof. Let L be a normal extension of K containing S , and let T be the integral closure of R in L. In particular T is integral over S , so we may choose Q2 ∈ Spec T lying over P2 and also Q1 ∈ Spec T lying over p1 . By the Going Up Theorem there exists Q′ ∈ Spec T containing Q1 and lying over p2 . Both Q2 and Q′ lie over p2 , so by Theorem 342 there exists σ ∈ Aut(L/K ) such that σ (Q′ ) = Q2 . Thus σ (Q1 ) ⊂ σ (Q′ ) = Q2 and σ (Q1 ) lies over p1 , so that setting P1 = σ (Q1 ) ∩ S we have P1 ∩ R = p1 and P1 ⊂ σ (Q′ ) ∩ S = Q2 ∩ S = P2 . Remark: In [?, Chapter 5] one finds a proof of Theorem 330 which avoids all Galoistheoretic considerations. However it is significantly longer than the given proofs of Theorems 342 and 330 combined and – to me at least – rather opaque. 14.5. The Noether Normalization Theorem. 14.5.1. The classic version. Theorem 331. (Noether Normalization I) Let k be a field, R = k [x1 , . . . , xm ] an integral domain which is a finitely generated k -algebra,51 and let K be the fraction 51Here the x ’s are not assumed to be independent indeterminates. i 196 PETE L. CLARK field of R. a) There exists d ∈ Z, 0 ≤ d ≤ m, and algebraically independent elements y1 , . . . , yd ∈ R such that R is finitely generated as a module over the polynomial ring k [y1 , . . . , yd ] – or equivalently, that R/k [y1 , . . . , yd ] is an integral extension. b) The integer d is equal to both the Krull dimension of R and the transcendence degree of K/k . Proof. a) (Jacobson) The result is trivial if m = d, so we may suppose m > d. Then the yi are algebraically dependent over k : there exists a nonzero polynomial ∑ f (s1 , . . . , sm ) = aJ sj1 · · · sjm , aJ ∈ k [s1 , . . . , sm ] m 1 with f (x1 , . . . , xm ) = 0. Let X be the set of monomials sJ = sj1 · · · sjm occuring m 1 in f with nonzero coefficients. To each such monomial we associate the univariate polynomial j1 + j2 t + . . . + jm tm−1 ∈ Z[t]. The polynomials obtained in this way from the elements of X are distinct. Since a univariate polynomial over a field has only finitely many zeroes, it follows that there exists a ≥ 0 such that the integers j1 + j2 a + . . . + jm am−1 obtained from the monomials in X are distinct. Now consider the polynomial m− 1 f (s1 , sa + t1 , . . . , sa 1 1 We have m−1 a f (s1 , sa + t1 , . . . , s1 1 = ∑ + tm ) = ∑ + tm ) ∈ k [s, t]. m− 1 aJ sj1 (sd + t2 )j2 · · · (sa 1 1 1 + tm )jm J aJ sj1 +j2 a+...+jm a 1 m−1 + g (s1 , t2 , . . . , ym ), J m m−1 ∑ in which the degree of g in s1 is less than that of J aJ sj1 +j2 a+...+j a . Hence 1 m−1 + tm ) is a monic polynomial in x1 for suitable β ∈ k × , βf (s1 , sa + t2 , . . . , sa 1 1 i− 1 with k [t2 , . . . , tm ]-coefficients. Putting wi = xi − xa for 2 ≤ i ≤ m, we get 1 m− 1 βf (x1 , xd + w2 , . . . , xa 1 1 + wm ) = 0, so that x1 is integral over R′ = k [w2 , . . . , wm ]. By induction on the number of generators, R′ has a transcendence base {yi }d=1 such that R′ is integral over k [y1 , . . . , yd ]. i Thus R is integral over k [y1 , . . . , yd ] by transitivity of integrality. b) Since R/k [y1 , . . . , yd ] is an integral extension, by Corollary 326 the Krull dimension of R is equal to the Krull dimension of k [y1 , . . . , yd ], which by Corollary 285 is d. Moreover, since R is finitely generated as a k [y1 , . . . , yd ], by Proposition 319 the fraction field K is finitely generated as a k (y1 , . . . , yd )-module, so the transcendence degree of K/k is equal to the transcendence degree of k (y1 , . . . , yd )/k , i.e., d. 14.5.2. Noether normalization over a domain. Theorem 332. (Noether Normalization II) Let R ⊂ S be domains with S finitely generated as an R-algebra. There exists a ∈ R• and y1 , . . . , yd ∈ S algebraically independent over the fraction field of R such that Sa (the localization of S at the multiplicative subset generated by a) is finitely generated as a module over T = Ra [y1 , . . . , yd ]. COMMUTATIVE ALGEBRA 197 Proof. (K.M. Sampath) Let K be the fraction field of R and let x1 , . . . , xm be a set of R-algebra generators for S . Then S ′ := S ⊗R K = K [x1 , . . . , xm ] is finitely generated over K (as above, the xi ’s need not be algebraically independent). Applying Theorem 331, we get algebraically independent elements y1 , . . . , yd ∈ S ′ such that S ′ is a finitely generated T ′ := K [y1 , . . . , yd ]-module. Multiplying by a suitable element of R× , we may assume yi ∈ S for all i. Since S ′ is finitely generated as a T ′ -module, it is integral over T ′ . For 1 ≤ i ≤ m, xi satisfies a monic polynomial equation with coefficients in T ′ : n n y1 + Pi,1 (y1 , . . . , yd )yi −1 + . . . + Pi,n = 0. Let a be the product of the denominators of all coefficients of all the polynomials Pi,k . It follows that Sa is integral and finitely generated as a T = Ra [y1 , . . . , yd ]algebra, hence it is finitely generated as a T -module. Exercise: In the setting of Theorem 332 suppose that S is a graded R-algebra. Show that we may take all the yi to be homogeneous elements. 14.5.3. Applications. The Noether Normalization Theorem is one of the foundational results in algebraic geometry: geometrically, it says that every integral affine variety of dimension d is a finite covering of affine d-space Ad . Thus it allows us to study arbitrary varieties in terms of rational varieties via branched covering maps. It is almost as important as a theorem of pure algebra, as even the “soft” part of the result, that the Krull dimension of an integral affine k -algebra is equal to the transcendence degree of its fraction field, is basic and useful. One of the traditional applications of Noether Normalization is to prove Hilbert’s Nullstellensatz. As we have seen, it is fruitful to channel proofs of the Nullstellensatz through Zariski’s Lemma, and this is no exception. Proposition 333. Noether Normalization implies Zariski’s Lemma. Exercise: Prove Proposition 333. Theorem 334. Let k be a field, and let R be a domain which is finitely generated as a k -algebra, with fraction field K . Then: a) The Krull dimension of R is equal to the transcendence degree of K/k . b) Every maximal chain of prime ideals in R has length dim R. Proof. a) By Noether normalization, R is finite over k [t1 , . . . , td ], so K is finite over k (t1 , . . . , td ). Thus the transcendence degree of K/k is d. On the other hand, R is integral over k [t1 , . . . , td ] so by Theorem X.X dim R = dim k [t1 , . . . , td ]. By Corollary XXX, dim k [t1 , . . . , td ] = d. b) See Madapusi-Sampath, p. 119... 14.6. A Little Invariant Theory. Let R be a commutative ring, let G be a finite group, and suppose G acts on R by automorphisms, i.e., we have a homomorphism ρ : G → Aut(R). We define RG = {x ∈ R |∀g ∈ G, gx = x}, 198 PETE L. CLARK the ring of G-invariants – it is indeed a subring of R. Remark: As in the case of rings acting on commutative groups, we say that Gaction on R is faithful if the induced homomorphism ρ : G → Aut(R) is injective. Any G-action induces a faithful action of G/ ker(ρ), so it is no real loss of generality to restrict to faithful G-actions. We will do so when convenient and in such a situation identify G with its isomorphic image in Aut R. The simplest case is that in which R = K is a field. Then K G is again a field and K/K G is a finite Galois extension. Conversely, for any finite Galois extension K/F , F = K Aut(K/F ) . This characterization of Galois extensions was used by E. Artin as the foundation for an especially elegant development of Galois theory (which swiftly became the standard one). Note also the analogy to topology: we have the notion of a finite Galois covering Y → X of topological spaces as one for which the group G = Aut(Y /X ) of deck transformations acts freely and properly discontinuously on Y such that Y /G = X . The branch of mathematics that deals with invariant rings under finite group actions is called classical invariant theory. Historically it was developed along with basic commutative algebra and basic algebraic geometry in the early 20th century, particularly by Hilbert. Especially, Hilbert’s work on the finite generation of invariant rings was tied up with his work on the Basis Theorem. For a ∈ R, define NG (a) = have defined a map ∏ σ ∈G σ (a). Observe that NG (a) ∈ RG , so that we NG : R → R G . Note that NG is not a homomorphism of additive groups. However, when R is a domain, there is an induced map NG : R• → (RG )• which is a homomorphism of monoids, so induces a homomorphism on unit groups. Exercise X.X.: Let R[t] be the univariate polynomial ring over R. Show that there is a unique action of G by automorphisms of G on R[t] extending the G-action on R and such that gt = t. Show that (R[t])G = RG [t]. Proposition 335. For a finite group G acting on R, R/RG is integral. Proof. For x ∈ R, define Φx (t) = NG (t − x) = ∏ (t − gx), g ∈G so Φx (t) ∈ (R[t])G = RG [t]. Thus Φx (t) is a monic polynomial with RG -coefficients which is satisfied by x. Base extension: Suppose that G is a finite group acting faithfully on R. Moreover, let A be a ring and f : A → R be a ring homomorphism, so R is an A-algebra. Suppose moreover that f (A) ⊂ RG . In such a situation we say that G acts on R by A-automorphisms and write G ⊂ Aut(R/A). COMMUTATIVE ALGEBRA 199 Suppose we have another A-algebra A′ . We can define an action of G on R ⊗A A′ by putting g (x ⊗ y ) := gx ⊗ y . We say that the G-action is extended to A′ . Proposition 336. In the above setup, suppose that A′ is a flat A-algebra. Then there is a natural isomorphism ∼ RG ⊗A A′ → (R ⊗A A′ )G . Proof. Madapusi p. 65-66. Corollary 337. Let G be a finite group acting on the ring R, and let S ⊂ RG be a multiplicatively closed set. Then (S −1 R)G = S −1 RG . Exercise: Prove Corollary 337. In particular, suppose R is a domain with fraction field K , and let F be the fraction field of RG . Then the G-action on R extends to a G-action on K , and Corollary 337 gives K G = F . Thus the invariant theory of integral domains is compatible with the Galois theory of the fraction fields. Proposition 338. If R is integrally closed, so is RG . Proof. Let x ∈ K be an element which is integral over RG . Then x is certainly also integral over R, and since R is integally closed in L we have x ∈ R. Thus x ∈ R ∩ K = R ∩ LG = RG . Theorem 339. (Noether [No26]) Suppose that R is a finitely generated algebra over some field k with k = k G . Then: a) R is finitely generated as an RG -module. b) RG is a finitely generated k -algebra. Proof. a) Since R is a finitely generated k -algebra and k ⊂ RG , R is a finitely generated RG -algebra. But by Proposition 335 R/RG is integral. So R is finitely generated as an RG -module. b) By part a), the Artin-Tate Lemma (Theorem 231) applies to the tower of rings k ⊂ RG ⊂ R. The conclusion is as desired: RG is a finitely generated k -algebra. Remark: The title of [No26] mentions “characteristic p”. In fact, when k has characteristic 0 the result had been proven by Hilbert significantly earlier [Hi90], and moreover for certain actions of infinite linear groups, like SLn (k ). But Noether’s formulation and proof give an excellent illustration of the economy and power of the commutative algebraic perspective. Let us now make contact with the setup of “classical invariant theory”. Let k be a field, V a finite-dimensional vector space and ρ : G → Autk (V ) a linear representation of G on V . Let k [V ] = Sym(V ∨ ) be the algebra of polynomial functions on V . If we choose a k -basis e1 , . . . , en of V and let x1 , . . . , xn be the dual basis of V ∨ , then k [V ] = k [x1 , . . . , xn ] is a polynomial ring in n independent indeterminates. There is an induced action of G on k [V ], namely for f ∈ k [V ] we put (gf )(x) = f (g −1 x). All of our above results apply in this situation. Especially, Theorem 339 applies to tell us that the ring k [V ]G is finitely generated as a k -algebra, or a finite system of invariants. Of course, we did not so much as crease our sleeves, let alone 200 PETE L. CLARK roll them up, to establish this: for a concretely given finite group G and action on a k -vector space V , it is of interest to explicitly compute such a finite system. Moreover, the polynomial ring k [V ] is integrally closed: in the next section we will see that it is a unique factorization domain and that this is a stronger property. Therefore Proposition 338 applies to show that k [V ]G is integrally closed. This is actually quite a robust and useful procedure for producing integrally closed rings. Example (the one example everyone knows): Let k be any field, n ∈ Z+ , let V = k n , let G = Sn be the symmetric group, and let G act on V by permuting the standard basis elements e1 , . . . , en . The induced action of G on k [V ] = k [x1 , . . . , xn ] is (up to an inversion, or alternately a choice of left versus right action) the one obtained by permuting the variables xi . An element of k [V ]G is called a symmetric polynomial or symmetric function. There are evident symmetric functions s1 , . . . , sn , ∑ namely s1 (x) = x1 + . . . + xn , s2 (x) = i<j xi xj , and so forth: sk is a sum of (n) the k monomials of degree k . Then the fundamental theorem on symmetric functions is that k [V ]G is a polynomial ring on the sk , i.e., k [V ]G = k [s1 , . . . , sn ], and the sk ’s are independent indeterminates.52 The above example is well-known and extremely useful, but gives a misleadingly simple impression of classical invariant theory. One can ask how often the ring of invariants of a finite group action on a polynomial ring is again a polynomial ring, and there is a nice answer to this. But let’s back up a step and go back to “rational invariant theory”: if G acts on k [x1 , . . . , xn ], then as above it also acts on the fraction field k (x1 , . . . , xn ) and we know that k (x1 , . . . , xn )/k (x1 , . . . , xn )G is a finite Galois extension. But must k (x1 , . . . , xn )G itself be a rational function field, as it was in the example above? This is known as Noether’s Problem: it was first posed by E. Noether in 1913. It is natural and important, for an affirmative answer would allow us to realize every finite group as a Galois group (i.e., the automorphism group of a Galois extension) of Q thanks to a famous theorem of Hilbert. For more than half of the twentieth century, Noether’s problem remained open. Finally, in 1969 R.G. Swan (yes, the same Swan as before!) found a representation of the cyclic group of order 47 on a finite-dimensional Q-vector space for which the invariant field is not a rational function field [Sw69]. Too bad – this was arguably the best shot that anyone has ever taken at the Inverse Galois Problem over Q.53 Example: Let k be a field of characteristic different from 2, let V = k 2 , and consider the action of the two-element group G = {±1} on V by −1 acting as the scalar matrix −1. The induced action on k [V ] = k [x, y ] takes x → −x and y → −y . This is, apparently, a not very interesting representation of a not very ineresting group. But the invariant theory is very interesting! Exercise: a) Show that k [V ]G is generated as a k -algebra by x2 , y 2 and xy . b) Show that k [V ]G is isomorphic to the k -algebra k [A, B, C ]/(AB − C 2 ). c) Show that nevertheless the fraction field of k [V ]G is rational, i.e., is isomorphic 52This will not be proved here. Someday it will appear in my field theory notes, but not today. 53Actually, Serre’s Topics in Galois Theory describes a conjecture of J.-L. Colliot-Th´l`ne – ee roughly a weaker form of Noether’s problem – which would still imply that every finite group is a Galois group over Q. I am not aware of any progress on this conjecture. COMMUTATIVE ALGEBRA 201 to k (X, Y ) for independent indeterminates X and Y . Before signing off on our quick glimpse of classical invariant theory, we cannot resist mentioning one more classic theorem in the subject. It answers the question: when is the invariant subalgebra k [V ]G isomorphic to a polynomial algebra over k ? Let ρ : G → GL(V ) be a faithful representation of G on a finite-dimensional k -vector space V . An element g ∈ GL(V ) is a pseudoreflection if it has finite order and pointwise fixes a hyperplane W in V . (Equivalently, a pseudoreflection has characteristic polynomial (t − 1)dim V −1 (t − ζ ), where ζ is a root of unity in k .) Exercise X.X: If k = R, any nontrivial pseudoreflection has order 2 – i.e., it really is a hyperplane reflection. A faithful representation ρ of G is a pseudoreflection representation of G if ρ(G) is generated by pseudorflections. Theorem 340. (Shephard-Todd-Chevalley) Let k be a field, and let G be a finite group with #G not divisible by the characteristic of k . Let ρ : G → GL(V ) be a faithful representation of G on a finite-dimensional vector space. TFAE: (i) k [V ]G is a polynomial algebra. (ii) k [V ] is a free k [V ]G -module. (iii) ρ is a pseudoreflection representation of G. A proof in the case k = C is given in [Neus07, Chapter 13]. Exercise X.X: It follows from Theorem 340 the fundamental theorem on symmetric functions that the standard permutation representation of the symmetric group Sn on k n is a pseudoreflection representation. Show this directly. 14.7. Abstract algebraic number theory I: Galois extensions of integrally closed domains. Proposition 341. Let G be a finite group acting by automorphisms on a ring R, with invariant subring RG . Let ι : RG → R, and let p ∈ Spec RG . a) There is a natural action of G on the fiber (ι∗ )−1 (p) – i.e., on the set of primes P of R such that ι∗ P = p. b) The G-action on the fiber (ι∗ )−1 (p) is transitive. Proof. Let P ∈ Spec R and σ ∈ G. Define σ P = {σx | x ∈ P}. It is straightforward to verify that σ P is a prime ideal of R (if you like, this follows from the fact that Spec is a functor). Moreover (σ P ) ∩ RG is the set of all elements σx with x ∈ P such that for all g ∈ G, gσx = σx. As g runs through all elements of G, so does gσ −1 , hence (σ P ) ∩ RG = P ∩ RG = p. b) Let P1 , P2 be two primes of R lying over a prime p of RG . Let x ∈ P1 . Then NG (x) ∈ P1 ∩ RG = p ⊂ P2 . Since P2 is prime, there exists at least one σ ∈ G ∪ such that σx ∈ P2 , and thus P1 ⊂ σ∈G σ P2 . By Prime Avoidance (Lemma 227), there exists σ ∈ G such that P1 ⊂ σ P2 . Since R/RG is integral, Incomparability (Corollary 324) yields P1 = σ P2 . 202 PETE L. CLARK Theorem 342. Let R be an integrally closed domain with fraction field K , let L/K be a normal algebraic field extension (possibly of infinite degree), and let S be the integral closure of R in L. Let p ∈ Spec R, and let Xp be the set of all prime ideals of S lying over p. Then G = Aut(L/K ) acts transitively on Sp . Proof. Step 0: Do Exercise X.X to see that the action of G on Xp is well-defined. Step 1: Suppose [L : K ] = n < ∞, and write G = {σ1 = 1, . . . , σr }.54 Seeking a − contradiction, suppose there are P1 , P2 ∈ Xp such that P2 ̸= σj 1 P1 for all j . By −1 Corollary 324, P2 is not contained in any σj P1 , so that by Lemma 227 (Prime ∪− Avoidance) there exists x ∈ P2 \ j σj 1 P1 . Let q be the inseparable degree of (∏ )q L/K and put y = j σj (x) . Thus y = NL/K (x), so y ∈ K . Moreover y is integral over the integrally closed domain R, so y ∈ R. But since σ1 = 1, y ∈ P2 , so y ∈ P2 ∩ R = p ⊂ P1 , and thus, since P1 is prime, σj (x) ∈ P1 for some j : contradiction! Step 2: We will reduce to the case in which L/K is a Galois extension. Let G = Aut(L/K ) and K ′ = LG , so that L/K ′ is Galois and K ′ /K is purely inseparable. Let R′ be the integral closure of R in K ′ . Then by Lemma 329 Spec R′ → Spec R is a bijection. So we may as well assume that K ′ = K and L/K is Galois. Step 3: For each finite Galois subextension M of L/K , consider the subset F (M ) := {σ ∈ G | σ (P1 ∩ M ) = P2 ∩ M }. Observe that F (M ) is a union of cosets of Gal(L/M ) hence is (open and) closed in the∏ Krull topology. By Step 1, we have F (M ) ̸= ∅. Moreover, the compositum M M = i Mi of any finite number {∩ i } of finite Galois subextenions is again a finite Galois subextension, and we have i F (Mi ) ⊃ F (M ) ̸= ∅. Therefore as Mi ranges through all finite Galois subextensions of L/K , {F (Mi )}i∈I is a family of closed subsets of the compact space G satisfying the finite intersection condition, and it ∩ follows that there exists σ ∈ i F (Mi ) = F (L) i.e., σ ∈ G such that σ P1 = P2 . 15. Factorization Let R be an integral domain, and x a nonzero, nonunit element of R. We say that x is irreducible if for any y, z ∈ R such that x = yz , one of y or z is a unit. Remarks: For any unit u ∈ R× , we get factorizations of the form x = u · (u−1 x), so every nonzero nonunit has at least these factorizations, which we wish to regard as “trivial.” On the other hand, y and z cannot both be units, for then x would also be a unit. Let us define a factorization of a nonzero nonunit a ∈ R as a product a = x1 · · · xn , such that each xi is irreducible. We say that two factorizations a = x1 · · · xn = y1 · · · ym are equivalent if the multisets of associated principal ideals {{(xi )}} = {{(yj )}} are equal. More concretely, this means that m = n and that there is a bijection 54Recall that we are assuming that L/K is normal, so L/K is separable iff L/K is Galois iff r = n. COMMUTATIVE ALGEBRA 203 σ : {1, . . . , n} → {1, . . . , m} such that (yσ(i) ) = (xi ) for all 1 ≤ i ≤ n. If factorizations always exist and any two factorizations of a given element are equivalent, we say R is a unique factorization domain (UFD). 15.1. Kaplansky’s Theorem (II). A basic and important result that ought to get covered at the undergraduate level is that PID implies UFD. In fact this is easy to prove. What is more difficult is to get a sense of exactly how UFDs are a more general class of rings than PIDs. In this regard, an elegant theorem of Kaplansky seems enlightening. Exercise X.X: Let x be an element of an integral domain which can be expressed as x = p1 · · · pn , such that for 1 ≤ i ≤ n, Pi = (pi ) is a prime ideal. If then there exist principal prime ideals Q1 , . . . , Qm such that (x) = Q1 · · · Qm , then m = n and there exists a permutation σ of the integers from 1 to n such that Qi = Pσ(i) for all i. Exercise X.X: Let R be a domain, and let S be the set of all nonzero elements x in R such that (x) can be expressed as a product of principal prime ideals. Show that S is a saturated multiplicatively closed subset. Theorem 343. (Kaplansky) An integral domain is a UFD iff every nonzero prime ideal in R contains a prime element. Proof. Suppose R is a UFD and 0 ̸= P is a prime ideal. Let x ∈ P be a nonzero nonunit. Write x = p1 · · · pr a product of prime elements. Then x ∈ P implies pi ∈ P for some i, so (pi ) ⊂ P . Conversely, assume each nonzero prime ideal of R contains a principal prime. Let S be the set of all products of prime elements, so that by Exercise X.X, S is a saturated multiplicative subset. By Exercise X.X, it is enough to show that S contains all nonzero nonunits of R. Suppose for a contradiction that there exists a nonzero nonunit x ∈ R \ S . The saturation of S implies S ∩ (x) = ∅, and then by Theorem 117 there exists a prime ideal P containing x and disjoint from S . But by hypothesis, P contains a prime element p, contradicting its disjointness from S. We deduce immediately: Corollary 344. Let R be a domain. a) If every ideal of R is principal, R is a UFD. b) Conversely, if R is a UFD of dimension one, every ideal of R is principal. Proof. a) Recall that a height one prime is a prime ideal p of R which does not strictly contain any nonzero prime ideal. Applying Kaplansky’s theorem, we get a prime element x such that 0 (x) ⊂ p, and height one forces equality. For part b), it clearly suffices to show that a Noetherian domain in which each height one prime 204 PETE L. CLARK is principal is a UFD. But let p be any nonzero prime ideal; since R is Noetherian, (DCC) holds on prime ideals, so p contains a prime ideal which has height one, hence contains a prime element and we are done by Kaplansky’s theorem. Remark: Applying b) to R = k [x1 , . . . , xn ], we get the important geometric fact that each irreducible codimension one subvariety in affine space is cut out by a single equation. 15.2. Atomic domains, (ACCP). If every nonzero nonunit admits at least one factorization, we say R is an atomic domain55. Exercise X.X: Show that the ring Z of all algebraic integers is not an atomic domain. Indeed, since for every algebraic integer x, there exists an algebraic integer y such that y 2 = x, there are no irreducible elements in Z! The condition of factorization into irreducibles (in at least one way) holds in every Noetherian domain. In fact, a much weaker condition than Noetherianity suffices: Proposition 345. Let R be a domain in which every ascending chain of principal ideals stabilizes. Then every nonzero nonunit factors into a product of irreducible elements. In particular, a Noetherian domain is atomic. Proof. Let R be a domain satisfying the ascending chain condition for principal ideals (ACCP for short), and suppose for a contradiction that R is not an atomic domain. Then the set of principal ideals generated by unfactorable elements is nonempty, so by our assumption there exists a maximal such element, say I = (a). Evidently a is not irreducible, so we can begin to factor a: a = xy where x and y are nonunits. But this means precisely that both principal ideals (x) and (y ) properly contain (a), so that by the assumed maximality of (a), we can factor both x and y into irreducibles: x = x1 · · · xm , y = y1 · · · yn . But then a = x1 · · · xm y1 · · · yn is a factorization of a, contradiction. This proposition motivates us to consider also the class of domains which satisfy the ascending chain condition for principal ideals (ACCP). Exercise X.X: Suppose that R → S is an exension of rings such that S × ∩ R = R× . (In particular, this holds for integral extensions.) Show that S satisfies (ACCP) implies R satisfies (ACCP). Does the converse hold? We have just seen that (ACCP) implies atomicity. The proof shows that under (ACCP) we can always obtain an expression of a given nonzero nonunit by a finite sequence of “binary factorizations” i.e., replacing an element x with y1 · y2 , where y1 and y2 are nonunits whose product is x. After a bit of thought, one is inclined to worry that it may be possible that this factorization procedure fails but nevertheless irreducible factorizations exist. This worry turns out to be justified: Theorem 346. There exists an atomic domain which does not satisfy (ACCP). 55In the first pass on these notes, I used the sensible name “factorization domain.” But apparently atomic domain is what is used in the literature, and since we find this terminology unobjectionable, we might as well use it here. COMMUTATIVE ALGEBRA 205 Proof. See [Gr74]. However, the following strenghtening of atomicity does imply ACCP: A domain R is a bounded factorization domain BFD if it is atomic and for each nonzero nonunit a ∈ R, there exists a positive integer N (a) such that in any irreducible factorization a = x1 · · · xr we have r ≤ N (a). Proposition 347. A UFD is a BFD. Proof. An immediate consequence of the definitions. Proposition 348. A BFD satisfies (ACCP). Proof. Let R be a BFD. Suppose for a contradiction that (xi )i∈Z+ is a strictly ascending chain of principal ideals. We therefore have x0 = y1 x1 = y1 y2 x2 = . . . = y1 · · · yn xn = · · · , with each xi , yi a nonunit. Since R is atomic, we can refine each factorization into an irreducible factorization, but clearly an irreducible refinement of y1 · · · yn xn has at least n + 1 irreducible factors, contradicting BFD. 15.3. EL-domains. An element x of a domain R is prime if the principal ideal (x) is a prime ideal. Equivalently, x satisfies Euclid’s Lemma: if x | yz , then x | y or x | z . Proposition 349. A prime element is irreducible. Proof. If x is reducible, then x = yz with neither y nor z a unit, so that yz ∈ (x) but y ̸ ∈(x), z ̸ ∈(x). However, it need not be the case that irreducible elements are prime! √ √ Example: Let R = Z[ −5]. Then the elements 2, 3 and 1 ± −5 are all irreducible, but the equation √ √ 2 · 3 = (1 + −5)(1 − −5) shows that none of them are prime elements. √ Exercise X.X: Check √ these assertions. Hint: Define N (a + b √−5) = a2 + 5b2 . all √ √ Check that N ((a + b −5)(c + d −5)) = N (a + b −5)N (c + d −5). Show that √ α | β (in R) =⇒ N (α) | N (β ) (in Z), and use this to show that 2, 3, 1 ± −5 are irreducible but not prime. It is tempting to call a domain in which all irreducible elements are prime “Euclidean”, but this terminology is already taken for domains satisfying a generalization of the Euclidean algorithm (see XX). So we will, provisionally, call a ring in which irreducible elements are prime an EL-domain. (EL = Euclid’s Lemma). Theorem 350. For an integral domain R, TFAE: (i) R is a UFD. (ii) R satisfies (ACCP) and is an EL-domain. (iii) R is an atomic EL-domain. 206 PETE L. CLARK Proof. i) =⇒ (ii): In the previous section we saw UFD =⇒ BFD =⇒ (ACCP). We show UFD implies EL-domain: let x ∈ R be irreducible and suppose x | yz . Let y = y1 · · · ym and z = z1 · · · zn be irreducible factorizations of y and z . Then the uniqueness of irreducible factorization means that x must be associate to some yi or to some zj , and hence x | y or x | z : R is an EL-domain. (ii) =⇒ (iii) follows immediately from Proposition 345. (iii) =⇒ (i): This is nothing else than the usual deduction of the fundamental theorem of arithmetic from Euclid’s Lemma: in a factorization domain we have at least one irreducible factorization of a given nonzero nonunit x. If we moreover assume that irreducibles are prime, this allows us to compare any two irreducible factorizations: suppose x = y1 · · · ym = z1 · · · zn . Then y1 is a prime element so divides zj for some j . WLOG, relabel to assume j = 1. Since z1 is irreducible, we have y1 = u1 z1 and thus we may cancel to get y2 · · · ym = (u−1 z2 )z3 · · · zn . 1 Continuing in this way we find that each yi is associate to some zj ; when we get ∏ down to 1 = j zj we must have no factors of zj left, so m = n and R is a UFD. 15.4. GCD-domains. For elements a and b of R, a greatest common divisor is an element d of R such that: d | a, d | b and for any e ∈ R such that e | a, e | b, then e | d. Exercise X.X: Show that if there exists a greatest common divisor of two elements a and b, then an element d′ of R is a gcd of a and b iff (d) = (d′ ). In particular, any two gcd’s are associate. If a and b have a gcd, it would be more logically sound to write gcd(a, b) to mean the unique principal ideal whose generators are the various gcd’s of a and b. It is traditional however to use the notation gcd(a, b) to denote an element, with the understanding that in general it is only well-defined up to multiplication by a unit. (In some rings, principal ideals have canonical generators: e.g. in the integers we may take the unique positive generator and in k [t] we may take the unique monic generator. Under these circumstances, one generally means gcd(a, b) to stand for this canonical generator.) More generally, for elements a1 , . . . , an in a domain R, a greatest common divisor is an element d of R such that d | ai for all i and if e | ai for all i then e | d. If a GCD of (a1 , . . . , an ) exists, it is unique up to associates, and we denote it by gcd(a1 , . . . , an ). As above, it can be characterized as the unique minimal principal ideal containing ⟨a1 , . . . , an . Moreover, these setwise GCDs can be reduced to pairwise GCD’s, e.g.: Exercise: Let a, b, c be elements of a domain R and assume that all pairwise GCD’s exist in R. Then gcd(a, b, c) exists and we have gcd(a, gcd(b, c)) = gcd(a, b, c) = gcd(gcd(a, b), c). A domain R is a GCD-domain if for all a, b ∈ R, gcd(a, b) exists. By the above COMMUTATIVE ALGEBRA 207 remarks, it would be equivalent to require that gcd(a1 , . . . , an ) for all n-tuples of elements in R. Proposition 351. (GCD identities) Let R be a GCD-domain. Then: a) For all a, b, c ∈ R, gcd(ab, ac) = a gcd(b, c). a b b) For all a, b ∈ R \ {0}, gcd( gcd(a,b) , gcd(a,b) ) = 1. c) For all a, b, c ∈ R, gcd(a, b) = gcd(a, c) = 1, then gcd(a, bc) = 1. d) For all a, b, c ∈ R, gcd(a, b + ac) = gcd(a, b). e) For all a, a1 , . . . , an , b1 , . . . , bn , c ∈ R, gcd(a, b1 +ca1 , . . . , bn +can ) = gcd(a, b1 , . . . , bn ). Proof. a) Let x = gcd(ab, ac). Then a | ab and a | ac so a |x: say ay = x. Since x | ab and x | ac, y | b and y | c, so y | gcd(b, c). If z | b and z | c, then az | ab and 1 az | ac, so az | x = ay and z | y . Therefore gcd(b, c) = y = a gcd(ab, ac). Part b) follows immediately. As for part c): suppose gcd(a, b) = gcd(a, c) = 1, and let t divide a and bc. Then t divides ab and bc so t | gcd(ab, bc) = b gcd(a, c) = b. So t divides gcd(a, b) = 1. d) If d divides both a and b, it divides both a and b + ac. If d divides both a and b + ac, it divides b + ac − c(a) = b. e) We have gcd(a, b1 + ca1 , . . . , bn + can ) = gcd(a, gcd(a, b1 + ac1 ), . . . , gcd(a, bn + acn )) = gcd(a, gcd(a, b1 ), . . . , gcd(a, bn )) = gcd(a, b1 , . . . , bn ). Proposition 352. A GCD-domain is an EL-domain. Proof. This follows from the fact gcd(x, y ) = gcd(x, z ) = 1 =⇒ gcd(x, yz ) = 1. Theorem 353. Consider the following conditions on a domain R: (i) R is a UFD. (ii) R is a GCD-domain. (iii) R is an EL-domain: irreducible elements are prime. Then (i) =⇒ (ii) =⇒ (iii) always; and (iii) =⇒ (i) if R satisfies (ACCP). In particular, a Noetherian domain is a UFD iff it is GCD iff it is EL. Proof. Assume (i), and let x and y be nonzero elements of R. We may write x = f1 · · · fr g1 · · · gs , y = uf1 · · · fr h1 · · · ht , where the f ’s, g ’s and h’s are prime elements, (gj ) ̸= (hk ) for all j, k and u ∈ R× . It is then easy to check that f1 · · · fr is a gcd for x and y . (ii) implies (iii) is Proposition 352. (iii) + (ACCP) implies (i) is Theorem 350. We can now present some simple results that are long overdue. An extremely useful fact in algebra is that any UFD is integrally closed in its fraction field. We give a slightly stronger result and then recall a classical application: Theorem 354. A GCD-domain is integrally closed. Proof. Let R be a gcd-domain with fraction field K . Suppose x ∈ K satisfies an equation xn + an−1 xn−1 + . . . + a1 x + a0 = 0, ai ∈ R. 208 PETE L. CLARK We may write x = s with s, t ∈ R, t ̸= 0. Proposition XXa) asserts that, after t dividing through by the gcd we may assume gcd(s, t) = 1. Substituting s for x and t clearing denominators as usual, we get ( ) sn = − an−1 tsn−1 + . . . + a1 tn−1 s + a0 tn , and we have the usual contradiction: t divides the right hand side, so t divides sn . But by Proposition XXc) gcd(s, t) = 1 implies gcd(sn , t) = 1. The only way out of a contradiction is that t is a unit, i.e., x ∈ R. Corollary 355. An algebraic integer which is a rational number is an integer: Z ∩ Q = Z. Exercise X.X: Prove Corollary 355. √ Thus e.g. one can derive the irrationality of 2: it is a root of the monic polynomial equation t2 − 2 = 0 but evidently not an integer, so cannot be rational. Proposition 356. (Compatibility of GCD’s with localization) Let R be a GCDdomain and S a multiplicative subset of R. Then: a) The localization S −1 R is again a GCD-domain. b) For all x, y ∈ R• , if d is a GCD for x and y in R, then it is also a GCD for x and y in S −1 R. Exercise X.X: Prove Proposition 356. 15.5. Polynomial rings over UFDs. Our goal in this section to show that if R is a UFD, then a polynomial ring in any number (possibly infinite) of indeterminates is again a UFD. This result generalizes the “undergraddy” fact that if k is a field, k [t] is a UFD; the corresponding fact that polynomials in k [t1 , . . . , tn ] factor uniquely into irreducibles is equally basic and important, and arguably underemphasized at the pre-graduate level (including high school, where factorizations of polynomials in at least two variables certainly do arise). If we can establish that R a UFD implies R[t] a UFD, then an evident induction argument using the “identity” R[t1 , . . . , tn , tn+1 ] = R[t1 , . . . , tn ][tn+1 ] gives us the result for polynomials in finitely many indeterminates over a UFD (which is, so far as I know, the most important case). It is then straightforward to deduce the case for arbitrary sets of indeterminates. There are several different ways to prove the univariate case. Probably the most famous is via Gauss’ Lemma. To give this argument, we need some preliminary terminology. Let R be any integral domain, and consider a nonzero polynomial f = an tn + . . . + a1 t + a0 ∈ R[t]. We say f is primitive if x ∈ R, x | ai for all i implies x ∈ R× . In a GCDdomain, this is equivalent to gcd(a1 , . . . , an ) = 1. In a PID, this is equivalent to ⟨a0 , . . . , an ⟩ = R. For a general domain, this latter condition is considerably stronger: e.g. the polynomial xt + y ∈ k [x, y ][t] is primitive but the coefficients do COMMUTATIVE ALGEBRA 209 not generate the unit ideal. Let us call this latter – usually too strong condition – naively primitive. Proposition 357. Let R be a domain, and f, g ∈ R[t] be naively primitive. Then f g is naively primitive. Proof. Suppose f and g are naively primitive but f g is not. Then by definition the ideal generated by the coeffiicents of f is proper, so lies in some maximal ideal m of R. For g ∈ R[t], write g for its image in the quotient ring (R/m)[t]. Then our assumptions give precisely that f , g ̸= 0 but f g = f g = 0. Thus f and g are zero divisors in the integral domain (R/m)[t], a contradiction. If R is a GCD-domain and 0 ̸= f ∈ R[t], we can define the content c(f ) of f to be the gcd of the coefficients of f , well-determined up to a unit. Thus a polynomial is primitive iff c(f ) = 1. Exercise X.X: Let R be a GCD-domain and 0 ̸= f ∈ R[t]. a) Show that f factors as c(f )f1 , where f1 is primitive. b) Let 0 ̸= a ∈ R. Show that c(af ) = ac(f ). Theorem 358. (Gauss’ Lemma) Let R be a GCD-domain. If f, g ∈ R[t] are nonzero polynomials, we have c(f g ) = c(f )c(g ). If we assume the stronger hypothesis that R is a UFD, we can give a very transparent proof along the lines of that of Proposition 357 above. Since this special case may be sufficient for the needs of many readers, we will give this simpler proof first, followed by the proof in the general case. Proof. (Classical proof for UFDs) The factorization f = c(f )f1 of Exercise X.X reduces us to the following special case: if f and g are primitive, then so is f g . Suppose that f g is not primitive, i.e., there exists a nonzero nonunit x which divides all of the coefficients of f g . Since R is a UFD, we may choose a prime element π | x. Now we may argue exactly as in the proof of Proposition 357: (R/(π )[t] is an integral domain, f and g are nonzero, but f g = f g = 0, a contradiction. And now the proof in the general case, which makes use of some GCD identities. Proof. (Haible) As above, we may assume that f = an tn + . . . + a1 t + a0 , g = bm tm + . . . + b1 t + b0 ∈ R[t] are both primitive, and we wish to show that f g = cm+n tm+n + . . . + c1 t + c0 is primitive. We go by induction on n. Since a primitive polynomial of degree 0 is simply a unit in R, the cases m = 0 and n = 0 are both trivial; therefore the base case m + n = 0 is doubly so. So assume m, n > 0. By Proposition X.X(x), we have c(f g ) = gcd(cn+m , . . . , c0 ) = gcd(an bm , gcd(cn+m−1 , . . . , c0 )) | gcd(an , gcd(cn+m−1 , . . . , c0 ))·gcd(bm , gcd(cn+m−1 , . . . , c0 )). Now gcd(an , gcd(cn+m−1 , . . . , c0 )) = gcd(an , cn+m−1 , . . . , c0 ) = gcd(an , cn+m−1 − an bm−1 , . . . , cn − an b0 , cn−1 , . . . , c0 ) = gcd(an , c((f − an tn )g ). Our induction hypothesis gives c((f − an tn )g ) = c(f − an tn )c(g ) = c(f − an tn ), so gcd(an , cn+m−1 −an bm−1 , . . . , cn −an b0 , cn−1 , . . . , c0 ) = gcd(an , c(f −an tn )) = c(f ) = 1. 210 PETE L. CLARK Similarly we have gcd(bm , gcd(cn+m−1 , . . . , c0 )) = 1, so c(f g ) = 1. Corollary 359. Let R be a GCD-domain with fraction field K , and let f ∈ R[t] be a polynomial of positive degree. a) The following are equivalent: (i) f is irreducible in R[t]. (ii) f is primitive and irreducible in K [t]. b) The following are equivalent: (i) f is reducible in K [t]. (ii) There exist g, h ∈ R[t] such that deg(g ), deg(h) < deg(f ) and f = gh. Proof. a) Assume (i). Clearly an imprimitive polynomial in R[t] would be reducible in R[t], so f irreducible implies c(f ) = 1. Suppose f factors nontrivially in K [t], as f = gh, where both g, h ∈ K [t] and have smaller degree than f . By Exercise X.X, we may write g = c(g )g1 , h = c(h)h1 , with g1 , h1 primitive, and then f = c(g )c(h)g1 h1 . But then g1 , h1 , being primitive, lie in R[t], and c(f ) = c(g )c(h) = c(gh) ∈ R, so the factorization takes place over R[t], contradiction. (ii) =⇒ (i) is similar but much simpler and left to the reader. b) That (ii) =⇒ (i) is obvious, so assume (i). Because we can factor out the content, it is no loss of generality to assume that f is primitive. Let f = g1 h1 with g1 , h1 ∈ K [t] and deg(g1 ), deg(h1 ) < deg(f ). Because R is a GCD-domain, we may ˜ g h write g = d˜ , h = d2 with g , h ∈ R[t] primitive. Then we have d1 d2 f = g h, and ˜˜ ˜˜ 1 × equating contents gives (d1 d2 ) = (1), so d1 , d2 ∈ R and thus the factorization f = gh has the properties we seek. Theorem 360. (Gauss) If R is a UFD, so is R[t]. We will give three different proofs of this important result, essentially due to Gauss. The first proof is via Gauss’s Lemma. Proof. Let K be the fraction field of R, and let f ∈ R[t]• . We know that K [t] is a PID hence a UFD, so we get a factorization f = cg1 · · · gr , with c ∈ R and each gi ∈ R[t] is primitive and irreducible. Then factoring c into irreducibles gives an irreducible factorization of f . If we had another irreducible factorization f = dh1 · · · hs , then unique factorization in K [t] gives that we have r = s and after permuting the factors have gi = ui hi for all i, where ui ∈ K × . Since both gi and hi are primitive, we must have ui ∈ R× , whence the uniqueness of the factorization. We note that this proof relies on knowing that K [t] is a UFD, which of course follows from the fact that polynomial division gives a Euclidean algorithm, as one learns in an undergraduate course. This is of course an adaptation of the proof that the ring Z is a UFD (the Fundamental Theorem of Arithmetic) essentially due to Euclid. But it is of interest to find alternate routes to basic and important results. Theorem 361. Let R be a domain with fraction field K . a) If R is an ACCP-domain, so is R[t]. b) If R is a GCD-domain, so is R[t]. c) Thus, once again, if R is a UFD, so is R[t]. COMMUTATIVE ALGEBRA 211 Proof. a) In an infinite ascending chain {(Pi )} of principal ideals of R[t], deg Pi is a descending chain of non-negative integers, hence eventually constant. Therefore for sufficiently large n we have Pn = an Pn+1 with an ∈ R and (an+1 ) ⊃ (an ). Since R is an ACCP domain, we have (an ) = (an+1 ) for sufficiently large n, hence also (Pn ) = (Pn+1 ) for sufficiently large n. b) (Haible, [Hai94]) Let f, g ∈ R[t]. We may assume that f g ̸= 0. As usual, write ˜ ˜ f = c(f )f and g = c(g )˜. Since K [t] is a PID, may take the gcd of f and g in g ˜ × ˜ ˜ K [t], say d. The choice of d is unique only up to an element of K , so by choosing ˜ the unit appropriately we may assume that d lies in R[t] and is primitive. We put ˜. d = gcd(c(f ), c(g )))d ˜ ˜ ˜ Step 1: We claim that d is a gcd of f and g in R[t]. Since d | f in K [t], we ˜ ˜ f ˜ ˜ may write d = a q with a, b ∈ R \ {0} and q ∈ R[t] primitive. Since bf = ad, we ˜ b b × ˜ ˜ ˜˜ have (b) = c(bf ) = c(ad) = (a), i.e., a ∈ R and thus d | f in R[t]. Similarly ˜ | g . Moreover, sice d ∈ f K [t] + g K [t], there exist u, v ∈ R[t] and c ∈ R \ {0} ˜ ˜ d˜ ˜ ˜ ˜ ˜ ˜ with cd = uf + v g . Suppose h ∈ R[t] divides both f and g . Then h | cd, and ˜ ˜ ˜) = (1). Writing cd = a q with q ∈ R[t] primitive, and equating contents c(h) | c(f h b ˜ d ˜ ˜ in bcd = ahq , we get (bc) = (a), hence h = ab q ∈ R[t], so h | d. c Step 2: We claim that d is a gcd of f and g in R[t]. Certainly we have ˜ ˜ (d) = (gcd(c(f ), c(g )d) | (c(f )f ) = (f ), ˜ so d | f . Similarly d | g . Conversely, let h ∈ R[t] divide f and g . Write h = c(h)h ˜ ∈ R[t] primitive. From h | f it follows that c(h) | c(f ) and thus h | f . ˜˜ for h ˜˜ ˜˜ Similarly h | f so h | g . Thus c(h) | gcd(c(f ), c(g )), h | d and thus finally h | d. c) If R is a GCD domain and an ACCP domain, it is also an atomic EL domain, hence a UFD by Theorem 350. Lindemann [Li33] and Zermelo [Ze34] (independently) gave (similar) striking proofs of the Fundamental Theorem of Arithmetic avoiding all lemmas and packaging the Euclidean division into a single inductive argument. Later several authors have recorded analogous proofs of Gauss’s Theorem (Theorem 360): the earliest instance we are aware of in the literature is due to S. Borofsky [Bor50]. We give a third, “lemmaless” proof of Theorem 360 here. Proof. It suffices to show that R[t] is an ACCP domain and an EL-domain. By Theorem 361a), R[t] is an ACCP domain. Now, seeking a contradiction, we suppose that R is an EL-domain but R[t] is not. Among the set of all elements in R[t] admitting inequivalent irreducible factorizations, let p be one of minimal degree. We may assume p = f1 · · · fr = g1 · · · gs , where for all i, j , (fi ) ̸= (gj ) and m = deg f1 ≥ deg f2 ≥ . . . ≥ deg fr , n = deg g1 ≥ deg g2 ≥ . . . ≥ deg gs , with n ≥ m > 0. Suppose the leading coefficient of f1 (resp. g1 ) is a (resp. b). Put q = ap−bf1 xn−m g2 · · · gs = f1 (af2 · · · fr −bxn−m g2 · · · gs ) = (ag1 −bf1 xn−m )g2 · · · gs . Thus q = 0 implies ag1 = bf1 xn−m . If, however, q ̸= 0, then deg(ag1 − bf1 xn−m ) < deg g1 , 212 PETE L. CLARK hence deg q < deg p and q has a unique factorization into irreducibles, certainly including g2 , · · · , gs and f1 . But then f1 must be a factor of ag1 − bf1 xn−m and thus also of ag1 . Either way ag1 = f1 h for some h ∈ R[t]. Since a is constant and f1 is irreducible, this implies h = ah2 , so ag1 = f1 ah2 , or g1 = f1 h2 , contradiction. Corollary 362. Let R be a UFD and let {ti }i∈I be any set of indeterminates. Then S = R[{ti }] is a UFD. Proof. The case in which I is finite follows ∪ immediately from Theorem XX by induction. When I is infinite, we have S = J R[{tj }], as J ranges over all finite subsets of I : this is just because any given polynomial can only involve finitely many indeterminates. For f ∈ S , let J be such that f ∈ R[{tj }], so f = p1 · · · pr is a factorization into prime elements of R[{tj }]. Any two factorizations in S would themselves lie in some subalgebra involving finitely many determinates, so the factorization must be unique. Remark: In particular, a polynomial ring in infinitely many variables over a field is a standard example of a non-Noetherian UFD. We want to mention without proof two negative results. Theorem 363. a) (Roitman) There exists an integrally closed atomic domain R such that R[t] is not atomic. b) (Anderson-Quintero-Zafrullah) There exists an EL domain R such that R[t] is not an EL domain. 15.6. Application: Eisenstein’s Criterion. The most famous criterion for irreducibility of univariate polynomials is named after Ferdinand Eisenstein, who used it in an 1850 Crelle paper. In fact, the version for polynomials over Z was apparently first given by T. Sch¨nemann in 1846 o (also in Crelle ). Nowadays it is common to state and prove a version of Eisenstein’s criterion with respect to a prime ideal in a UFD. We give a slight generalization: Theorem 364. (Sch¨nemann-Eisenstein Criterion) Let R be a domain with fraco tion field K , and let f (t) = ad td + . . . + a1 t + a0 ∈ R[t]. Suppose that there exists a prime ideal p of R such that ad ̸∈ p, ai ∈ p for all 0 ≤ i < d and a0 ̸∈ p2 . a) If f is primitive, then f is irreducible over R[t]. b) If R is a GCD-domain, then f is irreducible over K [t]. Proof. a) Suppose to the contrary that f is primitive and reducible over R[t]: i.e., there exists a factorization f = gh with g (t) = bm tm + . . . + b1 t + b0 , h(t) = cn tm + . . . + c1 t + c0 , deg(g ), deg(h) < deg(f ) and bm cn ̸= 0. Since a0 = b0 c0 ∈ p \ p2 , it follows that exactly one of b0 , c0 lies in p: say it is c0 and not b0 . Moreover, since ad = bm cn ̸∈ p, cn ̸∈ p. Let k be the least index such that ck ̸∈ p, so 0 < k ≤ n. Then b0 ck = ak − (b1 ck−1 + . . . + bk c0 ) ∈ p. Since p is prime, it follows that at least one of b0 , ck lies in p, a contradiction. b) Suppose R is a GCD-domain and (seeking a contradiction) that f is reducible over K [t]. By Corollary 359b), we may write f = gh with g, h ∈ R[t] and deg(g ), deg(h) < deg(f ). Then the proof of part a) goes gives a contradiction. Corollary 365. Let R be a GCD-domain containing a prime element π (e.g. a UFD that is not a field). Then the fraction field K of R is not separably closed. COMMUTATIVE ALGEBRA 213 Proof. To say that π is a prime element is to say that the principal ideal p = (π ) is a nonzero prime ideal. Then π ̸∈ p2 , so for all n > 1, Pn (t) = tn − π is Eisenstein with respect to p and hence irreducible in K [t]. Choosing n to be prime to the characteristic of K yields a degree n separable field extension Ln := K [t]/(Pn ). 15.7. Application: Determination of Spec R[t] for a PID R. Let R be a PID. We wish to determine all prime ideals of the ring R[t]. Let us begin with some general structural considerations. First, R is a one-dimensional Noetherian UFD; so by Theorems 218, 361 and 229, R[t] is a two-dimensional Noetherian UFD. Being a UFD, its height one ideals are all principal. Since it has dimension two, every nonprincipal prime ideal is maximal. Therefore it comes down to finding all the maximal ideals. However, to avoid assuming the Dimension Theorem, we begin with milder hypotheses on a prime ideal P , following [R, pp. 22-23]. Namely, let P be a nonzero prime ideal of R[t]. We assume – only! – that P is not principal. By Theorem 343, we are entitled to a prime element f1 of P . Since P = (f1 ), let f2 ∈ P \ (f1 ). Then gcd(f1 , f2 ) = 1: since gcd(f1 , f2 ) | f1 , the only ̸ other possibility is (gcd(f1 , f2 )) = (f1 ), so f1 | f2 and f2 ∈ (f1 ), contradiction. First Claim: Let K be the fraction field of R. The elements f1 and f2 are also relatively prime in the GCD-domain K [t]. Indeed, suppose that f1 = hg1 , f2 = hg2 with h, g1 , g2 ∈ K [t] and h a nonunit. By Gauss’ Lemma, we may write h = ah0 , g1 = b1 γ1 , g2 = b2 γ2 with a1 , b1 , b2 ∈ K and h0 , γ1 , γ2 primitive elements of R[t]. Again by Gauss’ Lemma, h0 γ1 and h0 γ2 are also primitive, so f1 = hg1 = (ab1 )(h0 γ1 ) ∈ R[t], which implies that ab1 ∈ R. Similarly, ab2 ∈ R, so h0 is a nonunit of R[t] which divides both f1 and f2 , contradiction. Let M := ⟨f1 , f2 ⟩, and put m = M∩ R. It remains to show that, as the notation suggests, M is a maximal ideal of R[t] and m is a maximal ideal of R. Second Claim: m ̸= 0. Since K [t] is a PID and f1 , f2 are relatively prime in K [t], there exist a, b ∈ K [t] such that af1 + bf2 = 1. Let 0 ̸= c ∈ R be an element which is divisible by the denominator of each coefficient of a and b: then (ca)f1 + (cb)f2 = c with ca, cb ∈ R, so that c ∈ m. Now put p = P ∩ R, so p = (p) is a prime ideal of the PID R. Moreover, p=P ∩R⊃M∩R=m 0, so p is maximal. Since P ⊃ p, P corresponds to a prime ideal in R[t]/pR[t] = (R/p)[t], a PID. Therefore P is generated by p ∈ p and an element f ∈ R[t] whose image in (R/p)[t] is irreducible. We therefore have proved: Theorem 366. Let R be a PID, and P ∈ Spec R[t]. Then exactly one of the following holds: (0) P has height 0: P = (0). (i) P has height one: P = (f ), for a prime element f ∈ R[t]. 214 PETE L. CLARK (ii) P has height two: P = ⟨p, f ⟩, where p is a prime element of R and f ∈ R[t] is an element whose image in (R/p)[t] is irreducible. Moreover both P and p := P ∩ R are maximal, and [R[t]/P : R/p] < ∞. Exercise: Suppose R has only finitely prime ideals, so is not a Hilbert-Jacobson ring. By Theorem 287, there is a maximal ideal m of R[t] such that m ∩ R = (0). Find one, and explain where m fits in to the classification of Theorem 366. 15.8. Characterization of UFDs among Noetherian domains. Theorem 367. For a Noetherian domain R, TFAE: (i) Every height one prime ideal of R is principal. (ii) R is a UFD. Proof. . . . 15.9. Power series rings over UFDs. Exercise X.X: Show that if R is ACCP, so is R[[t]]. In particular, if R is a UFD, R[[t]] is ACCP. But of course the more interesting question is the following: Must R[[t]] be a UFD? In contrast to Gauss’s Theorem, whether a formal power series ring over a UFD must be a UFD was a perplexing problem to 20th century algebraists and remained open for many years. Some special cases were known relatively early on. Theorem 368. (R¨ckert [R¨33], Krull [Kr37]) Let k be a field, and let n be a u u positive integer. Then k [[t1 , . . . , tn ]] is a UFD. A significant generalization was proved by Buchsbaum and Samuel, independently, in 1961. We define the height of a prime ideal p in a ring R to be the supremum of all non-negative integers n such that there exists a strictly ascending chain of prime ideals p0 p1 . . . pn = p. In a domain R, a prime ideal has height 0 iff it is the zero ideal. A Noetherian domain R is regular if for every maximal ideal m of R, the height of m is equal to the dimension of m/m2 as a vector space over the field R/m. Theorem 369. ([Buc61], [Sa61]) If R is a regular UFD, then so is R[[t]]. In the same 1961 paper, Samuel also provided the first example of a UFD R for which R[[t]] is not a UFD [Sa61, §4]. One may also ask about formal power series in countably infinitely many indeterminates. Let k be a field. There is more than one reasonable way to define such a domain. One the one hand, one could simply take the “union” (formally, direct limit) of finite formal power series rings k [[t1 , . . . , tn ]] under the evident inclusion maps. In any element of this ring, only finitely many indeterminates appear. However, it is useful also to consider a larger ring, in which the elements are infinite formal k -linear combinations of monomials ti1 · · · tin . Let us call this latter domain k [[t1 , . . . , tn , . . .]]. In fact, we have seen this domain before: it is isomorphic to the Dirichlet ring Dk of functions f : Z+ → k with pointwise addition and convolution product. To see this, we use unique factorization in Z! Namely, write enumerate the prime numbers COMMUTATIVE ALGEBRA 215 ∏∞ {pi }∞ and write n ∈ Z+ as n = i=1 pai , where ai ∈ Z+ and ai = 0 for all i=1 i sufficiently large i. Then the map which sends f ∈ Dk to the formal power series ∑ ∏∞ ai n∈Z+ f (n) i=1 ti gives an isomorphism from Dk to k [[t1 , . . . , tn , . . .]]. In 1959, E.D. Cashwell and C.J. Everett used Theorem 368 to prove the following result. A key part of their proof was later simplified by C.F. Martin, who pointed out the applicability of K¨nig’s Infinity Lemma. o Theorem 370. ([CaEv59], [Mar71]) a) For any field k , the ring of formal power series k [[t1 , . . . , tn , . . .]] is a UFD. b) In particular, the ring DC = {f : Z+ → C} of airthmetic functions is a UFD. In almost any first number theory course one studies unique factorization and also arithmetic functions, including the Dirichlet ring structure (which e.g. leads to an immediate proof of the M¨bius Inversion Formula). That arithmetic functions are o themselves an example of unique factorization is however a very striking result that does not seem to be well-known to most students or practitioners of number theory. I must confess, however, that as a working number theorist I know of no particular application of Theorem 370. I would be interested to learn of one! 15.10. Nagata’s Criterion for UFDs. Proposition 371. Let R be a domain, S a saturated multiplicative subset, and f ∈ R \ S . If f is prime as an element of R, it is also prime as an element of RS . Proof. Since f ∈ R \ S , f is not a unit in RS . Let α, β ∈ RS be such that f | αβ 1 2 3 in RS . So there exists γ ∈ RS such that γf = αβ ; putting α = x1 , β = x2 , γ = x3 s s s and clearing denominators, we get s1 s2 x3 f = s3 x1 x2 , so f | r3 x1 x2 . If f | s3 , then since S is saturated, f ∈ S , contradiction. So, being prime, f divides x1 or x2 in 1 2 R, hence a fortiori in RS and therefore it also divides either x1 or x2 in RS , since s s these are associates to x1 and x2 . Theorem 372. Every localization of a UFD is again a UFD. Exercise: Prove Theorem 372. (Suggestions: one gets an easy proof by combining Theorem 343 with Proposition 371. But the result is also rather straightforward to prove directly.) A saturated multiplicative subset S of R is primal56 if it is generated by the units of R and by the prime elements of S . Lemma 373. An irreducible element of a primal subset is prime. Proof. Suppose S is primal and f ∈ S is irreducible. By definition, there exists a unit u and prime elements π1 , . . . , πn such that f = uπ1 · · · πn . Since uπ1 is also prime, we may as well assume that u = 1. Then, since f is irreducible, we must have n = 1 and f = π1 . Theorem 374. For an atomic domain R, the following are equivalent: (i) Every saturated multiplicative subset of R is primal. (ii) R is a UFD. 56This terminology is my invention: do you like it? 216 PETE L. CLARK Proof. Since the set R× of units is trivially generated by the empty set of prime elements, both conditions hold if R is a field, so let us now assume otherwise. Assume (i). Then, since R is a factorization domain which is not a field, there exists an irreducible element f of R. Let S be the saturated multiplicative subset generated by S , which consists of all units of R together with all divisors of positive powers f n of f . Since S is primal and strictly contains R× , there must exist a prime element π which divides f n for some n. In other words, f n ∈ πR, and since πR is prime, we must have that f = xπ for some x ∈ R. Since f is irreducible we must have x ∈ R× , i.e., f ∼ π and is therefore a prime element. So R is an ACCP domain and an EL-domain and hence a factorization domain by Theorem ??. Assume (ii), let S be a saturated multiplicative subset of R, and suppose that a a f ∈ S \ R× . Then f = uπ1 1 · · · πnn where the πi ’s are prime elements. Since each πi | f , πi ∈ S for all i. It follows that indeed S is generated by its prime elements together with the units of R. Because of Theorem 374, it is no loss of generality to restate Theorem 372 as: the localization of a UFD at a primal subset is again a UFD. The following elegant result of Nagata may be viewed as a converse. Theorem 375. (Nagata [Nag57]) Let R be a factorization domain and S ⊂ R a primal subset. If the localized domain RS is a UFD, then so is R. Proof. By Theorem 350 it suffices to show that if f ∈ R is irreducible, f is prime. Case 1: f ̸∈ S , so f is not a unit in RS . Since RS is a UFD, it is enough to show 1 2 that f is irreducible in RS . So assume not: f = x1 · x2 with x1 , x2 ∈ R \ S and s s s1 , s2 ∈ S . hen s1 s2 f = x1 x2 . By assumption, we may write s1 = up1 · · · pm and s2 = vq1 · · · qn , where u, v ∈ R× and pi , qj are all prime elements of R. So p1 | x1 x2 ; 1 2 since p1 is a prime, we must have either x1 ∈ R or x2 ∈ R. Similarly for all the p q other pi ’s and qj ’s, so that we can at each stage divide either the first or the second factor on the right hand side by each prime element on the left hand side, without 1 12 leaving the ring R. Therefore we may write f = ( uv ) x1 x2 where t1 , t2 are each tt products of the primes pi and qj , hence elements of S , and also such that t1 | x1 , t2 | x2 , i.e., the factorization takes place in R. Moreover, since xi ∈ R \ S and ti ∈ S , xii is not even a unit in RS , hence a fortiori not a unit in R. Therefore we t have exhibited a nontrivial factorization of f in R, contradiction. Case 2: f ∈ S . Since S is primal, by Lemma 373, f is prime. Remark: If S is the saturation of a finitely generated multiplicative set, the hypothesis that R is a factorization domain can be omitted. Application: Let A be a UFD and consider the polynomial ring R = A[t]. Put S = A \ {0}. As for any multiplicative subset of a UFD, S is generated by prime elements. But moreover, since A[t]/(πA[t]) ∼ (A/πA)[t], every prime element π of = A remains prime in A[t], so viewing S as the multiplicative subset of A[t] consisting of nonzero constant polynomials, it too is generated by prime elements. But if F is the fraction field of A, RS = (A[t])S = F [t] which is a PID and hence a UFD. Nagata’s theorem applied to R and S now tells us – for the third time! – that R = A[t] is a UFD. Nagata used Theorem 375 to study the coordinate rings of affine quadric cones. COMMUTATIVE ALGEBRA 217 Let k be a field of characteristic different from 2, and let f (x) = f (x1 , . . . , xn ) ∈ k [x1 , . . . , xn ] be a quadratic form, i.e., a homogeneous polynomial of degree 2 with k coefficients. We assume that f the associated bilinear form (x, y ) → 1 2 (f (x + y ) − f (x) − f (y )) is nonsingular. Equivalently, by making an invertible linear change of variables every quadratic form can be diagonalized, and a quadratic form is nonsingular iff it admits a diagonalization (19) f (x) = a1 x2 + . . . + an x2 witha1 , . . . , an ∈ k × . 1 n We wish to study the affine quadric cone associated to f , namely Rf = k [x]/(f ). Note that if quadratic forms f and g are isometric – i.e., differ by an invertible linear change of variables – then Rf ∼ Rg , so we assume if we like that f is in = diagonal form as in (19) above. If n ≥ 3 then every nonsingular diagonal quadratic polynomial is irreducible, so Rf is a domain. If k is quadratically closed – i.e., admits no proper quadatic extension – then conversely any binary (n = 2) quadratic form is reducible, so Rf is not a domain. (If f is not quadratically closed, there exist irreducible binary quadratic forms, but we will not consider them here.) Theorem 376. Let f = f (x1 , . . . , xn ) ∈ C[x1 , . . . , xn ] be a nonsingular quadratic form. Then Rf = k [x]/(f ) is a UFD iff n ≥ 5. Proof. By the remarks above, Rf is a domain iff n ≥ 3, so we may certainly restrict to this case. Because C is algebraically closed, every quadratic form in n ≥ 2 variables is isotropic, i.e., there exists 0 ̸= a ∈ k n such that f (a) = 0: indeed, the first n − 1 coordinates of a may be chosen arbitrarily. By an elementary theorem in the algebraic theory of quadratic forms [Lam06, Thm. I.3.4], we may make a change of variables to bring f into the form: f (x) = x1 x2 + g (x3 , . . . , xn ). Case 1: Suppose n = 3, so that 2 f (x) = x1 x2 − ax3 for some a ∈ k × . In this case, to show that Rf is not a UFD it suffices to show that the images x1 , x2 , x3 of x1 , x2 , x3 in Rf are nonassociate irreducibles, for then x1 x2 = ax3 2 exhibits a non-unique factorization! To establish this, regard k [x1 , x2 , x3 ] as a graded C-algebra in the usual way – with x1 , x2 , x3 each of degree 1 – so that the quotient Rf by the homogeneous ideal (f ) inherits a grading. Since x1 has degree 1, if it were reducible, it would factor as the product of a degree one element c1 x1 + c2 x2 + x3 x3 + (f ) and a degree zero element r + (f ), and thus (rc1 − 1)x1 + rc2 x2 + rc3 x3 ∈ (f ). But the left hand side has degree 1, whereas all nonzero elements in (f ) have degree 2 or higher, so r ∈ C[x]× and therefore the factorization is trivial. The irreducibility of x2 and x3 is proved in the same way. If x1 ∼ x3 in Rf , then we may divide both sides of x1 x2 − ax3 2 by x1 and deduce that also x2 ∼ x3 . But in the quotient ring Rf /(x3 ), x3 maps to 0 and x1 and x2 do not, contradiction. So Rf is not a UFD. Case 2: Suppose n = 4, so f (x) = x1 x2 + g (x3 , x4 ), where g (x3 , x4 ) is a nonsingular binary form. Here for the first time we use the full strength of the quadratic closure of k : since k × = k ×2 , any two nonsingular quadratic forms in the same number of variables are isometric, so we may assume WLOG that f (x) = x1 x2 − x3 x4 . 218 PETE L. CLARK Now we argue exactly as in Case 1 above: in Rf , the images x1 , x2 , x3 , x4 are all non-associate irredcuble elements, so x1 x2 = x3 x4 is a non-unique factorization. Case 3: n ≥ 5. Then n − 2 ≥ 3, so g is irreducible in the UFD C[x3 , . . . , xn ], hence also in C[x2 , x3 , . . . , xn ]. Therefore Rf /(x1 ) = C[x1 , . . . , xn ]/(x1 , f ) = C[x2 , . . . , xn ]/(g ) is a domain, i.e., x1 is a prime element. Moreover, R[x1 −1 ] = C[x1 , . . . , xn , x−1 ]/(x1 x2 − g ) 1 ∼ C[x1 , . . . , xn , x−1 ]/(x2 − g ) ∼ C[x1 , x3 , . . . , xn , x−1 ] = = 1 1 x1 is a localization of the UFD C[x1 , x3 , . . . , xn ] hence a UFD. By Nagata’s Criterion (Theorem 375), Rf itself is a UFD. Now let k be an arbitrary field of characteristic not 2 and f ∈ k [x1 , . . . , xn ] a nonsingular quadratic form. Without changing the isomorphism class of Rq we may diagonalize f ; moreover without changing the ideal (f ) we may scale by any element of k × , so without loss of generality we need only consider forms x2 +a2 x2 +. . .+an x2 . n 1 2 Theorem 377. Let k be a field of characteristic different from 2 and f = x2 + 1 a2 x2 + . . . + an x2 a nonsingular quadratic form over k . Put Rf = k [x]/(f ). n 2 a) If n ≤ 2 then Rf is not an integrally closed domain. b) If n = 3, Rf is a UFD iff f is anistropic: ∀a ∈ k n , f (a) = 0 =⇒ a = 0. c) (i) Suppose f = x2 − ax2 − bx2 − cx2 . If a is a square in k , then Rf is a UFD 4 3 2 1 iff −bc is not a square in k . (i) If none of a, b, c, −ab, −ac, −bc is a square in k , then Rf is a UFD iff −abc is not a square. d) If n ≥ 5, Rf is a UFD. Remark: If f (x1 , . . . , x4 ) s a diagonal quadratic form, we may permute and/or rescale the coefficients so that either condition (i) or condition (ii) of part c) holds. Proof. a) If n ≤ 2, Rf is never an integrally closed domain. b) The proof of Theorem 376 goes through to show that if f is isotropic (i.e., not anisotropic), Rf is not a UFD. The anisotropic case is due to Samuel [Sa64]. Part c) is due to T. Ogoma [O74]. Part d) goes back at least to van der Waerden [vdW39]. In [Nag57], M. Nagata gives a short proof using Theorem 375. It is also interesting to consider affine rings of inhomogeneous quadric hypersurfaces. For instance, we state without proof the following result. Theorem 378. For n ≥ 1, let Rn := R[t1 , . . . , tn+1 ]/(t2 + . . . + t2 +1 − 1) be the n 1 ring of polynomial functions on the n-sphere S n . a) (Bouvier [Bou78]) If n ≥ 2, then Rn is a UFD. b) (Trotter [T88]) R1 is isomorphic to the ring R[cos θ, sin θ] of real trigonometric polynomials, in which (sin θ)(sin θ) = (1+cos θ)(1 − cos θ)) is an explicit non-unique factorization into irreducible elements. Hence R1 is not a UFD. 15.11. Auslander-Buchsbaum. In the realm of algebraic geometry, the following is perhaps the single most important result about UFDs: Theorem 379. (Auslander-Buchsbaum) A regular local ring is a UFD. COMMUTATIVE ALGEBRA 219 A Noetherian local ring R with maximal ideal m is regular if the dimension of m/m2 as an R/m-vector space is equal to the Krull dimension of R. This is an algebraic expression of the geometric property of having a well-defined tangent space at the point m. If, for instance, R = C[x1 , . . . , xn ]/I is the coordinate ring of an affine complex variety, then maximal ideals correspond to points Pm of the zero set V (I ) ⊂ Cn , and regularity of the localization at m occurs iff the corresponding topological space has the structure of a smooth manifold in some neighborhood around the point Pm . ´ 16. Principal rings and Bezout domains A ring R in which every ideal is principal is called a principal ring. If R is moreover a domain, it is called a principal ideal domain (PID). 16.1. Principal ideal domains. We begin with the following result, essentially a restatement of X.X in §X.X . Proposition 380. a) A PID has dimension at most one. b) A UFD is a PID iff it has dimension at most one. Proof. a) Any domain has positive Krull dimension. Moreover, if (x) ⊂ (y ) are two nonzero principal prime ideals, then we may write x = cy for some 0 ̸= c ∈ R. Since x is prime, either x | c or x | y . If dx = c, then dcy = dx = c, and cancellation gives that y is a unit, contradiction. Therefore y = dx = dcy and cancellation gives dc = 1, i.e., c, d ∈ R× and (x) = (y ). b) We recall only that by Kaplansky’s Theorem 343, any nonzero ideal in a UFD contains a prime element. Exercise X.X: Let R be a PID and let S ⊂ R \ {0} be a multiplicative subset. Show that the localization S −1 R is a PID (which is a field iff S ∩ R× ̸= ∅). Theorem 381. A domain in which each prime ideal is principal is a PID. Proof. Suppose not. Then the set of all nonprincipal ideals is nonempty. Let {Ii } ∪ be a chain of nonprincipal ideals and put I = i Ii . If I = (x) then x ∈ Ii for some i, so I = (x) ⊂ Ii implies I = Ii is principal, contradiction. Therefore by Zorn’s Lemma there exists an ideal I which is maximal with respect to the property of not being principal. As is so often the case for ideals maximal with respect to some property or other, we can show that I is necessarily prime. Indeed, suppose that xy ∈ I but neither x nor y lies in I . Then the ideal J = ⟨I, x⟩ is strictly larger than I , hence principal: say J = (d). Let I ′′ = (I : x) be the set of all elements of r of R such that rx ∈ I . Then I ′′ is an ideal containing I and y , hence strictly larger than I and thus principal: say I ′′ = (c). Let r ∈ I , so r = ud. Now u(d) ⊂ I so ux ∈ I so u ∈ I ′′ . Thus we may write u = vc and r = vcd. This shows I ⊂ (cd). Conversely, c ∈ I ′′ implies cx ∈ I so c(I + xR) = cJ ⊂ I so cd ∈ I . Therefore I = (cd) is principal, contradiction. We now follow with some very familiar examples. Proposition 382. The integer ring Z is a principal ideal domain. 220 PETE L. CLARK To be honest, I can hardly imagine a reader without prior knowledge of this result (a staple of undergraduate mathematics) who has made it this far. Therefore, with our tongue slightly in cheek, we present a “structural” proof. To make clear the rules of our little game, let us say in advance that the only “prior knowledge” we will require about the ring Z is that for any positive integer n, the quotient Z/nZ has (finite!) order n. To show this, let z be any nonzero integer; note that the set of integers k such that z − kn ≥ 0 is bounded above so therefore has a largest element K ; and thus 0 ≤ z − Kn < n. Lemma 383. The integer ring Z is a unique factorization domain. Proof. Concretely, we claim that for every integer n > 1, there exist not necessarily distinct prime numbers p1 , ..., pr such that n = p1 · · · pr and also that if we have any s prime numbers q1 , . . . , qs such that n = q1 · · · qs , then r = s and there exists a permutation σ of {1, . . . , r} such that for all 1 ≤ i ≤ r we have qi = pσ(i) . Here are two proofs which the reader may not have seen before: First proof: Indeed a decomposition n = p1 · · · pr corresponds to a composition series for the Z-module Z/nZ. Since Z/nZ is finite, it is certainly Noetherian and Artinian, so composition series exist. Moreover the Jordan-H¨lder theorem o implies that any two composition series have the same number of terms – i.e., r = s = ℓ(Z/nZ) – and that after a permutation the sequences of isomorphism classes of composition factors become identical. Second proof (Lindemann [Li33], Zermelo [Ze34]): We prove both the existence and uniqueness of the factorization by an inductive argument, specifically by appeal to the well-ordering of the positive integers under ≤. Existence: let S be the set of integers n > 1 which do not have at least one prime factorization. We wish to show that S is empty so, seeking a contradiction, suppose not. Then by well-ordering S has a least element, say N . If N is prime, then we have found a prime factorization, so suppose it is not prime: that is, we may write N = N1 N2 with 1 < N1 , N2 < N . Thus N1 and N2 are too small to lie in S so each have prime factorizations, say N1 = p1 · · · pr , N2 = q1 · · · qs , and then N = p1 · · · pr q1 · · · qs gives a prime factorization of N , contradiction! Uniqueness: we claim that the factorization of a positive integer is unique. Assume not; then the set of positive integers which have at least two different standard form factorizations is nonempty, so has a least element, say N , where: (20) N = p1 · · · pr = q1 · · · qs . Here the pi ’s and qj ’s are prime numbers, not necessarily distinct from each other. However, we must have p1 ̸= qj for any j . Indeed, if we had such an equality, then we could cancel and, by an inductive argument we have already rehearsed, reduce to a situation in which the factorization must be unique. In particular p1 ̸= q1 . Without loss of generality, assume p1 < q1 . Then, if we subtract p1 q2 · · · qs from both sides of (20), we get (21) M := N − p1 q2 · · · qs = p1 (p2 · · · pr − q2 · · · qs ) = (q1 − p1 )(q2 · · · qs ). By the assumed minimality of N , the prime factorization of M must be unique. However, (21) gives two different factorizations of M , and we can use these to get a COMMUTATIVE ALGEBRA 221 contradiction. Specifically, M = p1 (p2 · · · pr − q2 · · · qs ) shows that p1 | M . Therefore, when we factor M = (q1 − p1 )(q2 · · · qs ) into primes, at least one of the prime factors must be p1 . But q2 , . . . , qj are already primes which are different from p1 , so the only way we could get a p1 factor is if p1 | (q1 − p1 ). But this implies p1 | q1 , and since q1 is also prime this implies p1 = q1 . Contradiction! Now we come to the proof of Proposition 382. Applying the lemma together with Kaplansky’s Theorem (Theorem 343), we see that every nonzero prime ideal p of Z contains a prime element, say p. But the quotient Z/(p) is a finite integral domain, hence a field, hence (p) is maximal and therefore (p) = p. This shows that every prime ideal of Z is principal, and now finally we apply Theorem 381. Proposition 384. For any field k , the ring R = k [t] is a principal ideal domain. Proof. By Theorem 361, R is a UFD. The remainder of the argument is quite similar to the case of R = Z. Namely it suffices to show that the ideal generated by any prime element is maximal. But a prime element of k [t] is, among other things, a polynomial p(t) of positive degree d, and the quotient R/(p(t)), being finite dimensional over a field k , is an Artinian integral domain. Thus R/(p(t)) is a field and (p(t)) is maximal. Conversely: Exercise X.X: Let R be a ring, and suppose that the univariate polynomial ring R[t] is a PID. Show that R is a field. Proposition 385. Let R be a Noetherian local ring with a principal maximal ideal m = (a). Then every nonzero ideal of R is of the form (ai ) for some i ∈ N. In particular, R is a principal ring. ∩ ∩ Proof. The key is the Krull intersection theorem: i mi = i (ai ) = 0. It follows that for any nonzero r ∈ R, there exists a largest i ∈ N such that r ∈ (ai ), i.e., there exists s ∈ R such that r = sai . But if s were not a unit then it would lie in m and thus r would lie in mi+1 , contradiction. So s is a unit and (r) = mi . Thus to every nonzero element r of I we attach a non-negative integer ir . Now if I is any nonzero ideal of R, choose a nonzero element r of I with ir minimal among elements of I . Then I ⊃ (r) = mi , and the other containment follows by minimality of ir . Corollary 386. For any field k , the formal power series ring k [[t]] is a PID. Proof. We have seen that a power series ring over a domain is a domain. By Theorem 218b), k [[t]] is Noetherian. Quite generally, for a ring R the units of R[[t]] are precisely those formal power series f = a0 + a1 t + . . . with a0 ∈ R× ; since k is a field this means that the units are those with nonzero constant term. The complement of the set of units is the principal ideal (t), so k [[t]] satisfies all the hypotheses of Proposition 385 and is therefore a PID. Remark: If we had available the theory of completions, we could deduce Corollary 386 directly from Proposition 384. 16.2. A word on Euclidean functions. 222 PETE L. CLARK We cannot quite allow ourselves to leave the topic of the previous section without making some mention of the standard way to show that rings like Z and k [t] are PIDS. Namely, if R is an integral domain, then a Euclidean function is a function N : R \ {0} → N such that: for all a ∈ R and b ∈ R \ {0}, there exists q, r ∈ R such that a = qb + r with r = 0 or N (r) < N (b). Exercise: For each of the following rings R, we give a function N : R \ {0} → N. Verify that N is a Euclidean function on R. a) R = Z, N (a) = |a|. b) R = k [t], N (p∑ = deg(p(t)). (t)) ∞ c) R = k [[t]], N ( n=0 an tn ) := the least n such that an ̸= 0. √ √ d) For d ∈ {−2, −1, 2, 3}, R = Z[ d] = Z[t]/(t2 − d), N (a + b d) = |a2 − db2 |. Exercise: Let R be a local PID. Use Proposition 385 to construct a Euclidean function on R. The virtue of Euclidean functions lies in the following result: Proposition 387. A domain R which admits a Euclidean function is a PID. Proof. Let I be any nonzero ideal of R, and choose an element x in I such that N (x) is minimal. For any y ∈ I , by definition of a Euclidean function we may write y = qx + r with r = 0 or N (y ) < N (x). Our choice of x rules out the second alternative and thus we have y = qx, i.e., y ∈ (x). Remark: It is common to define a Euclidean domain as a domain which admits some Euclidean function. In my opinion, this definition is somewhat of a false step in the theory. The merit of the notion of a Euclidean function is that for certain rings R there is an evident Euclidean function (e.g. for Z and k [t]), and when this occurs one deduces immediately that R is a PID. Moreover, if the Euclidean function is (in some sense) computable, the Euclidean algorithm can be used to find greatest common divisors and, accordingly, generators of ideals. In contrast, since the specific Euclidean function is not part of the definition of a Euclidean domain, in order to show that a given ring R is not Euclidean one needs to argue that no Euclidean function can exist. Evidently if R is not a PID, no such function can exist. However, there are PIDs which admit no Euclidean function (more details are given below). One way to assess the naturality of the notion Euclidean domain is to examine its stability under small perturbations of the definition. In other words, to what extent do similar axioms for a “Euclidean function” lead to an equivalent class of rings? There are several results to the effect that a certain weakening of the definition of a Euclidean function captures precisely the class of all PIDs. For example: Theorem 388. (Dedekind-Hasse) For a domain R with fraction field K , TFAE: (i) There exists a function N : K → Q satisfying: (QN1) ∀x ∈ K, N (x) ≥ 0; N (x) = 0 ⇐⇒ x = 0. (QN2) ∀x, y ∈ K, N (xy ) = N (x)N (y ). (QN3) ∀x ∈ R, N (x) ∈ Z. COMMUTATIVE ALGEBRA 223 (QN4) ∀x ∈ R, N (x) = 1 ⇐⇒ x ∈ R× . (QN5) ∀x ∈ K \ R, ∃ a, b ∈ R such that 0 < N (ax − b) < 1. (ii) R is a principal ideal domain. Exercise: Prove Theorem 388. (Suggestions: that (i) =⇒ (ii) is the usual argument that an ideal is generated by any element of minimal norm. Conversely, if R is a PID, define N on R \{0} by, e.g., setting N (x) to be 2r , where r = ℓ(R/xR) is the number of irreducibles appearing in a factorization of x and extend this map to K .) In another direction, P. Samuel considered the notion of a W-Euclidean function on a domain R. Here W is a well-ordered set and N : R → W is a function such that for all a ∈ R, b ∈ R \ {0} such that b a, ∃ q, r ∈ R with a = qb + r and N (r) < N (b). If R admits, for some W , a W -Euclidean function, then R is a PID. Exercise: Show that a domain is W -Euclidean for some finite W iff it is a field. Say that a domain is Samuel-Euclidean if it is W -Euclidean for some √ well-ordered set W . Samuel remarks that the an imaginary quadratic fields Q( −d) has a Samuel-Euclidean ring of integers Rd iff d = 1, 2, 3, 7, 11. On the other hand, it goes back at least to Gauss that for each of d = 19, 43, 67, 163 the ring Rd is a PID. Thus there are PIDs which are not Samuel-Euclidean. Samuel further showed that any Samuel-Euclidean ring is W -Euclidean for a unique minimal well-ordered set (up to canonical order isomorphism) WR and asked the question of whether one has WR ≤ N for all domains R. This was answered in the negative by Hiblot [Hi75]. 16.3. Some structure theory of principal rings. Proposition 389. Let R be a principal ring. a) For any multiplicative subset S , the localization RS is principal. b) For any ideal I of R, the quotient R/I is principal. Exercise X.X: Prove Proposition 389. Proposition 390. Let R1 , . . . , Rn be finitely many rings. TFAE: (i) Each Ri is a principal ring. ∏n (ii) The direct product i=1 Ri is a principal ring. Exercise X.X: a) Prove Proposition 390. b) Exhibit an infinite direct product of PIDs which is not a principal ring. Definition: A principal ring (R, m) is special if it is a local Artinian ring, i.e., if it is local and the maximal ideal is principal and nilpotent. The complete structure of ideals in special principal rings can be deduced from Proposition 385: if n is the least positive integer such that mn = 0, then the ideals of R are precisely the powers mi = (π i ) for 0 ≤ i ≤ n. We can now state a structure theorem for principal rings, which appears in Zariski and Samuel (so far as I know, for the first time). Theorem 391. (Zariski-Samuel) Every principal ring is the direct product of a finite number of PIDs and special principal rings. Much of the work of the proof is contained in the following preparatory result: 224 PETE L. CLARK Lemma 392. Let R be a principal ring. a) Let p q be prime ideals of R. Then q contains no prime ideals other than itself and p and every primary ideal contained in q contains p. b) If p is a non-maximal prime ideal and a ⊂ p is a primary ideal, then a = p. c) Any two incomparable prime ideals of R are comaximal. Proof. a) Suppose that we have p q and also q r, where r is a prime ideal. Write p = (p), q = (q ) and r = (r). There exists a ∈ R such that p = aq . Since p is a prime element and p does not divide q , we must have p | a, so that there exists b ∈ R with a = pb and thus p = pbq . Now since p(1 − bq ) = 0 ∈ r ⊂ q and 1 − bq ∈ R \ q, r does not contain 1 − bq and therefore we must have p ∈ r, i.e., p ⊂ r ⊂ q. But now modding out by p we get 0 ⊂ r/p ⊂ q/p in the principal ideal domain R/p, which is a one-dimensional ring, and therefore r = p or r = q. b) . . . c) Now let p1 and p2 be incomparable prime ideals. If p1 is maximal, then it is comaximal with any prime ideal which is not contained in it. Similarly, p2 is not maximal either. So if they are not comaximal, there exists a maximal ideal q strictly containing both p1 and p2 . But applying part a) to p1 q gives a contradiction. Proof of Theorem 391: Let (0) = n ∩ ai i=1 be an irredundant primary decomposition of (0), and let pi = rad(ai ). Then the pi ’s are pairwise comaximal; otherwise, for any distinct {i, j }, by Lemma 392 we would have, say, pi pj and then part b) of the lemma gives pi = ai ⊂ aj , contradicting irredundancy. Since the radicals of the ai ’s are pairwise comaximal, so are the ai ’s themselves (Proposition XX). So we may apply the Chinese Remainder Theorem, getting n n ∩ ∏ R = R/(0) = R/ ai = R/ai . i=1 i=1 Each factor is, of course, a principal ring. Fix any 1 ≤ i ≤ n. Case 1: Suppose first that pi is maximal. If ai were contained in any other prime ideal p, then rad(ai ) = pi ⊂ rad(p) = p, contradiction. So R/ai is a special PIR. Case 2: Otherwise, pi = ai , so that R/ai is a PID. We quote without proof the following somewhat stronger result: Theorem 393. (Hungerford structure theorem [Hun68]) For a commutative ring R, TFAE: (i) R is a principal ring. ∏n (ii) R ∼ i=1 Ri , and each Ri is a quotient of a PID. = Exercise: Derive Theorem 391 from Theorem 393. Exercise: Show that for a commutative ring R, TFAE: (i) R is a special principal ring. (ii) R is the quotient of a DVR by a nonzero ideal. Exercise: Let k be any field. Let R = k [x, y ], and m be the maximal ideal ⟨x, y ⟩ COMMUTATIVE ALGEBRA 225 in R. Show that the quotient ring S = R/m2 is nonprincipal. In particular, if k is finite, then S is a finite nonprincipal local ring. 16.4. B´zout domains. e Proposition 394. Let a, b be elements of a domain. If the ideal ⟨a, b⟩ is principal, then its generator is a greatest common divisor of a and b. Proof. In other words, we are assuming the existence of some d ∈ R such that dR = aR + bR. Then a, b ∈ dR, so d is a common divisor of a and b. If e | a and e | b then e ∈ aR + bR = dR, so e | d. Corollary 395. For a domain R, TFAE: (i) Every finitely generated ideal is principal. (ii) For any two elements a and b of R, gcd(a, b) exists and is an R-linear combination of a and b. Proof. (i) =⇒ (ii) is immediate from Proposition XX: gcd(a, b) will be a generator of the ideal ⟨a, b⟩. Conversely, if d = gcd(a, b) exists and is of the form d = xa + yb for some x, y ∈ R, then clearly (d) = ⟨a, b⟩, so that every ideal with two generators is principal. By an obvious induction argument, we conclude that any finitely generated ideal is principal. ´ At least according to some, it was Etienne B´zout who first explicitly noted that e for polynomials P, Q ∈ k [t], gcd(a, b) exists and is a linear combination of a and b: this fact is called B´zout’s identity or B´zout’s Lemma. For this (somee e what tenuous) reason, a possibly non-Noetherian domain satisfying the equivalent conditions of Corollary 395 is called a B´zout domain. e Theorem 396. For a B´zout domain R, the following are equivalent: e (i) R is a PID. (ii) R is Noetherian. (iii) R is a UFD. (iv) R is an ACCP domain. (v) R is an atomic domain. Proof. (i) ⇐⇒ (ii) immediately from the definitions. (i) =⇒ (iii): this is Corollary 344. (iii) =⇒ (iv) =⇒ (v) holds for all domains. (v) =⇒ (iii): A B´zout domain is a GCD-domain is an EL-domain, so a B´zout e e atomic domain is a UFD. (iv) =⇒ (ii): assume that R is not Noetherian. Then it admits an ideal I which is not finitely generated, which we can use to build an infinite strictly ascending chain of finitely generated ideals I1 I2 . . . I . But since R is B´zout, each Ii e is principal, contradicting ACCP. Say that a domain is properly B´zout if it is B´zout but not a PID. e e Recall that we have already seen some interesting examples of properly B´zout e domains, namely the ring of entire functions (Theorem 147) and the ring of all algebraic integers (Theorem 129). To see many more examples of B´zout domains (including properly B´zout doe e mains), we move on to the next topic: valuation rings. 226 PETE L. CLARK 17. Valuation rings 17.1. Basic theory. Consider the divisibility relation – i.e., a | b – on a domain R. Evidently it is reflexive and transitive, so is a quasi-ordering.57 Divisibility need not be a partial ordering because a | b and b | a does not imply that a = b but only that a and b are associates: (a) = (b). However, one of the first ideas of ideal theory is to view associate elements as being somehow “equivalent.” This motivates us to consider the equivalence relation on R in which a ∼ b iff (a) = (b). This is easily seen to be a monoidal equivalence relation. In plainer language, if (a1 ) = (a2 ) and (b1 ) = (b2 ), then (a1 b1 ) = (a2 b2 ). We can therefore consider the commutative monoid of principal ideals of R under multiplication, on which the divisibility relation is a partial ordering. Having made a quasi-ordering into a partial ordering, it may be natural to ask for conditions under which the divisibility relation induces a total ordering. Equivalently, for any a, b ∈ R either a | b or b | a. Proposition 397. Let R be a domain with fraction field K . TFAE: (i) For every a, b ∈ R, a | b or b | a. (ii) For every 0 ̸= x ∈ K , x ∈ R or x−1 ∈ R. Exercise: Prove Proposition X.X. A domain R satisfying the conditions of Proposition 397 is called a valuation domain or valuation ring. Note that any field is a valuation ring. This is a trivial example which is often implicitly excluded from consideration (we will try our best to be explicit in our exclusion of trivial cases). Apart from this, in a first algebra course one may not see examples of valuation rings. But we have: if p is a prime number, then the ring y Z(p) of integers localized at p is such an example. Define x |p y if ordp ( x ) ≥ 0. Then p-divisibility is immediately seen to be a total quasi-ordering: given two integers, at least one p-divides the other. The fundamental theorem of arithmetic implies x | y ⇐⇒ ∀ primes p, x |p y. However, in Z(p) , we have x | y ⇐⇒ x |p y , i.e., we have localized the divisibility relation to get a total quasi-order: Z(p) is a valuation domain. This argument generalizes as follows: let R be a PID58 and p = (π ) be a prime ideal of R. We define ordp (x) to be the least n such that (x) ⊃ pn , and extend it to a map on K × by ordp ( x ) = ordp (x) − ordp (y ). (One should check that this is y y well-defined; this is easy.) Finally, we define x |p y to mean ordp ( x ) ≥ 0. Arguing as above, we see that the localization Rp is a valuation ring. Note that in showing that Rp was a valuation domain we proceeded by constructing a map ordp on the nonzero elements of the fraction field K . This can be generalized, as follows: if R is a domain with quotient field K , we can extend the divisibility y y relation to K × by saying that x | y iff x ∈ R. Clearly x | y and y | x iff x is a unit 57By definition, a quasi-ordering is a reflexive, transitive binary relation on a set. 58In fact we can take R to be any Dedekind domain, as soon as we know what such a thing is. See §18. COMMUTATIVE ALGEBRA 227 in R. Therefore the quotient of (K × , ·) on which divisibility (from R!) becomes a partial ordering is precisely the quotient group K × /R× . y For [x], [y ] ∈ K × /R× , let us write [x] ≤ [y ] if [ x ] ∈ R. (Take a second and check that this is well-defined.) Exercise X.X: Show that the divisibility quasi-ordering on R is a total ordering iff the ordering on K × /R× is a total ordering. In other words, if R is a valuation ring, then the canonical map v : K × → K × /R× is a homomorphism onto a totally ordered abelian group. Let us relabel the quotient group by G and denote the group law by addition, so that the homomorphism property gets recorded as (VRK1) ∀x, y ∈ K × v (ab) = v (a) + v (b). We recover R as R = {x ∈ K × | v (x) ≥ 0} ∪ {0}. Everything that has been said so far takes into account only the multiplicative structure on R. So the following additional property is very important: (VRK2) ∀x, y ∈ K × | x + y ̸= 0, v (x + y ) ≥ min(v (x), v (y )). Indeed, suppose WLOG that v (x) ≤ v (y ), i.e., v (x) ≤ v (x + y ). y x ∈ R. Then x+y x = 1+ y x ∈ R so Exercise: Suppose that v (x) ̸= v (y ). Show that v (x + y ) = min(v (x), v (y )). Let (G, +, ≤) be a totally ordered abelian group. We write G+ = {g ∈ G | g ≥ 0}, so G+ is a totally ordered submonoid of G. A (G-valued) valuation on a field K is a surjective map v : K × → G satisfying (VRK1) and (VRK2) above. Exercise: Let v : K × → G be a valuation. Let R be the set of elements of K × with non-negative valuation, together with 0. Show that R is a valuation ring with fraction field K . Exercise: Let R be a domain, G a totally ordered group and v : R \ {0} → G+ be a map which satisfies both of the following properties: (VRR1) ∀x, y ∈ R \ {0}, v (ab) = v (a) + v (b). (VRR2) ∀x, y ∈ R \ {0} | x + y ̸= 0, v (x + y ) ≥ min(v (x), v (y )). (VRR3) v (R \ {0}) ⊃ G+ . Show that there is a unique extension of v to a valuation v : K × → G, namely v (x/y ) = v (x) − v (y ). Proposition 398. A valuation ring is a local domain, in which the unique maximal ideal m consists of elements of positive valuation. Proof. The set of elements of strictly positive valuation form an ideal m of R. Its complement, the set of elements of zero valuation, is the group of units. 228 PETE L. CLARK Proposition 399. Let I be a finitely generated ideal in a valuation ring. Then the set v (I ) has a least element, and for any x ∈ I of minimal valuation, I = v (x). Proof. Write I = ⟨x1 , . . . , xn ⟩ with v (x1 ) ≤ . . . ≤ v (xn ). Then for any r1 , . . . , rn ∈ R, v (r1 x1 + . . . + rn xn ) ≥ mini v (ri xi ) ≥ v (x1 ), so x1 is an element of I of minimal xi xi valuation. Thus for all i > 1, v ( x1 ) ≥ 0, so x1 ∈ R and x1 | xi . So I = ⟨x1 ⟩. Corollary 400. A valuation ring is a B´zout domain. In particular, a Noetherian e valuation ring is a PID. Conversely: Proposition 401. A local B´zout domain is a valuation domain. e Proof. Let x, y be elements of a local B´zout domain, and suppose d = gcd(x, y ). e Then x and y are coprime. Since in a local ring the nonunits form an ideal, this d d implies that at least one of x and y is a unit. In other words, up to associates, d d d = x (so x | y ) or d = y (so y | x). A valued field is a field together K together with a valuation v : K × → G. An isomorphism of valued fields (K, v ) → (K ′ , v ′ ) is an isomorphism f : K → K ′ of fields such that v = v ′ ◦ f . A valued field is trivial if G is the trivial group. Evidently each field carries a unique trivial valuation up to isomorphism: the valuation ring is K itself. The reader will lose nothing by making the tacit assumption that all valuations are nontrivial. Exercise: A chain ring is a ring R in which the partially ordered set I (R) of ideals of R is linearly ordered. a) Show that for a ring R, TFAE: (i) R is a chain ring. (ii) For all x, y ∈ R, either xR ⊂ yR or yR ⊂ xR. b) Show that a domain R is a chain ring iff it is a valuation ring. 17.2. Ordered abelian groups. Let (G, +) be an abelian group, written additively. In particular the identity element of G will be denoted by 0. As for rings, we write G• for G \ {0}. By an ordering on G we mean a total (a.k.a. linear) ordering ≤ on G which is compatible with the addition law in the following sense: (OAG) For all x1 , x2 , y1 , y2 ∈ G, x1 ≤ x2 and y1 ≤ y2 implies x1 + y1 ≤ x2 + y2 . One has the evident notions of a homomorphism of ordered abelian groups, namely an isotone group homomorphism. Exercise: Let (G, ≤) be an ordered abelian group. a) Let x ∈ G• . Show that either x > 0 or −x > 0 but not both. b) Show that for all x, y ∈ G, x ≤ y ⇐⇒ −y ≤ −x. Exercise: Let (G, ≤) be an ordered abelian group, and let H be a subgroup of G. Show that the induced order on H makes H into an ordered abelian group. COMMUTATIVE ALGEBRA 229 Example: For any ordered field (F, ≤), the additive group (F, +) is an ordered abelian group. In particular, the additive group (R, +) of the real numbers is an ordered abelian group, as is any subgroup. In particular, (Z, +) and (Q, +) are ordered abelian groups. Exercise: Exhibit an abelian group which admits two nonisomorphic orderings. Example (Lexicographic ordering): Let {Gi }i∈I be a nonempty indexed family of ordered abelian groups. Suppose that we are given a well-ordering on the index ∏ set I . We may then endow the direct product G = i∈I Gi with the structure of an ordered abelian group, as follows: for (gi ), (hi ) ∈ G, we decree (gi ) < (hi ) if for the least index i such that gi ̸= hi , gi < hi . Exercise: Check that the lexicographic ordering on the product a total ordering on G. ∏ i ∈I Gi is indeed Theorem 402. (Levi [Lev43]) For an abelian group G, TFAE: (i) G admits at least one ordering. (ii) G is torsionfree. Proof. (i) =⇒ (ii) Suppose < is an ordering on G and let x ∈ G• . Then exactly one of x, −x is positive; without loss of generality say it is x. Then for all n ∈ Z+ , nx = x + . . . x (n times) must be positive, so x has infinite order in G. (ii) =⇒ (i): Let G be a torsionfree abelian group. By Corollary 93, G is a flat Z-module. Tensoring the injection Z → Q gives us an injection G → G ⊗ Q. Since ⊕ Q is a field, the Q-module G ⊗ Q is free, i.e., it is isomorphic to i∈I Q. Choose a total ordering on I . Give each copy of Q its standard ordering as a subfield of R and ⊕ put the lexicographic ordering on i∈Q Q ∼ G ⊗ Q. Via the injection G → G ⊗ Q = this induces an ordering on G. Exercise X.X: a) Show that the abelian group Z admits exactly one ordering (here when we say “ordering”, we always mean “ordering compatible with the group structure in the sense of (OAG). b) Give an example of an abelian group which admits two distinct – even nonisomorphic – orderings. An ordered abelian group (G, +) is Archimedean if for all x, y ∈ G with x > 0, there exists n ∈ Z+ with nx > y . Exercise X.X: a) Suppose H is a subgroup of the Archimedean ordered group (G, +). Show that the induced ordering on G is Archimedean. b) Let (G, +) be an ordered abelian group such that there exists an embedding (G, +) → (R, +) into the additive group of the real numbers. Deduce that G is Archimedean. Conversely: o o Theorem 403. (H¨lder [H¨01]) Let (G, +) be an ordered abelian group. If G is Archimedean, there exists an embedding (G, +) → (R, +). 230 PETE L. CLARK Proof. We may assume G is nontrivial. Fix any positive element x of G. We will construct an order embedding of G into R mapping x to 1. Namely, let y ∈ G. Then the set of integers n such that nx ≤ y has a maximal element n0 . Put y1 = y − n0 x. Now let n1 be the largest integer n such that nx ≤ 10y1 : observe that 0 ≤ n1 < 10. Continuing in this way we get ∑ of integers a set ∞ n n1 , n2 , . . . ∈ {0, . . . , 9}. We define φ(y ) to be the real number n0 + k=1 10k . It is k ′ ′ not hard to show that φ is isotone – y ≤ y =⇒ φ(y ) ≤ φ(y ) – and also that φ is injective: we leave these tasks to the reader. But let us check that φ is a homomorphism of groups. For y ∈ G, and r ∈ Z+ , n let 10r be the rational number obtained by truncating φ(y ) at r decimal places. The numerator n is characterized by nx ≤ 10r y < (n + 1)x. For y ′ ∈ G, if n′ x ≤ 10r y ′ ≤ (n′ + 1)x, then (n + n′ )x ≤ 10r (y + y ′ ) < (n + n′ + 2)x, so φ(y + y ′ ) − (n + n′ )10−r < 2 10r and thus 4 . 10r Since r is arbitrary, we conclude φ(y + y ′ ) = φ(y ) + φ(y ′ ). |φ(y + y ′ ) − φ(y ) − φ(y ′ )| < A nontrivial ordered abelian group which can be embedded in R is said to have rank one. This is indeed part of a general definition of the rank of an ordered abelian group (so that, e.g., the lexicographic product of n copies of Z has rank n), but as we have no need for the general concept and the definition is not immediately perspicuous, we omit it here. The following result gives another characterization of valuation rings of rank one. Proposition 404. For a valuation ring R, TFAE: (i) R has rank one, i.e., the value group is Archimedean. (ii) R has Krull dimension one. Proof. (i) =⇒ (ii): Since the value group is nontrivial, R is not a field. Let p be a prime ideal which is strictly contained in the maximal ideal m of R. Choose α ∈ m \ p and put x = v (α). Let β ∈ p• and put y = v (β ). Using the Archimedean n property of the value group, there exists n ∈ Z+ such that nx > y and thus α ∈ R. β It follows that αn ∈ βR ⊂ p, and since p is prime, α ∈ p, contradiction. (ii) =⇒ (i): Let β ∈ m• . Since m is the unique prime containing β , r(βR) = m: for all α ∈ m there is n ∈ Z+ with β | αn . So the value group is Archimedean. Lemma 405. Let R be a domain, (G, ≤) a totally ordered commutative group, and let G+ = {g ∈ G | g ≥ 0}. Then: a) G+ is an ordered commutative monoid. b) The monoid ring R[G+ ] is an integral domain. c) The group ring R[G] is naturally isomorphic to the localization of R[G+ ] at the multiplicative subset G+ . In particular, R[G] is an integral domain. Exercise X.X: a) Write out the statements of Lemma 405 when G = Z. b) Prove Lemma 405. COMMUTATIVE ALGEBRA 231 Theorem 406. (Malcev, Neumann) For any ordered abelian group G, there exists a valuation domain with value group isomorphic to G. Proof. It suffices to construct a field K and a surjective map v : K × → G satisfying (VD1K) and (VD2K). Let k be any field and put A = k [G≥0 ]. By Lemma 405, A is a domain; let ∑ be its fraction field. Define a map v : A• → G by sending a K nonzero element g∈G ag g to the least g for which ag ̸= 0. It is easy to check that v satisfies (VD1)-(VD3) of Exercise X.X above, and therefore extends uniquely to a valuation v : K × → G, where K is the fraction field of R. Recall from §X.X the notion of a “big monoid ring” k [[Γ]], the collection of all functions f : Γ → k under pointwise addition and convolution product. As we saw though, in order for the convolution product to be defined “purely algebraically” – i.e., without recourse to some limiting process – we need to impose the condition of divisor finiteness on Γ. It follows easily from Proposition 411 that for Γ = G≥0 the monoid of non-negative elements in a totally ordered abelian group, divisor finiteness holds iff G = Z, i.e., iff the valuation is discrete. However, Malcev [Mal48] and Neumann [Neum49] independently found a way around this by considering a set in between k [G≥0 ] and k [[G≥0 ]]. Namely, define kMN [G≥0 ] to be the set of all functions f : G≥0 → k such that the support of f – i.e., the set of g ∈ G≥0 such that f (g ) ̸= 0 – is well-ordered. It turns out that on such functions the convolution product can be defined and endows kMN [G≥0 ] with an integral domain. The fraction field kMN (G) is simply the collection of all functions f : G → k with well-ordered support. Moreover, mapping each such nonzero function to the least element of G in its support gives a G-valued valuation. The elements of such rings are called Malcev-Neumann series. 17.3. Connections with integral closure. Let (R, mR ) and (T, mT ) be local rings with R ⊂ T . We say that T dominates R, and write R ≤ T , if mT ∩ R = mR . Lemma 407. Let R be a subring of a field K , and let p ∈ Spec R. Then there exists a valuation ring T of K such that R ⊂ T and mT ∩ R = p. Proof. (Matsumura) Step 0: We may replace R by Rp and thus asume that (R, p) is a local ring. In this case, what we are trying to show is precisely that there exists a valuation ring of K dominating R. Step 1: Let F be the set of all rings R′ with R ⊂ R′ ⊂ K such that pR′ R′ , partially ordered by inclusion. We have R ∈ F , so F ̸= ∅. Moreover the union of a chain in F is again an element of F , so Zorn’s Lemma gives us a maximal element T of F . Since pT T , there exists a maximal ideal m of T containing pT . Since T ⊂ Tm and Tm ∈ F , by maximality of T we have T = Tm , so (T, m) is a local ring dominating (Rp , p). Step 2: We claim that T is a valuation ring. proof of claim: Let x ∈ K × . We wish to show that at least one of x, x−1 lies in T . Seeking a contradiction, assume neither is the case. Then T [x] properly contains T , so by maximality of T we have 1 ∈ pT [x], i.e., we get a relation of the form 1 = a0 + a1 x + . . . + an xn , ai ∈ pT. 232 PETE L. CLARK Since T is local, 1 − a0 ∈ T × , and the relation may be rewritten in the form (22) 1 = b1 x + . . . + bn xn , bi ∈ m. Among all such relations, we may choose one with minimal exponent n. In exactly the same way, T [x−1 ] properly contains T and thus there exists a relation (23) 1 = c1 x−1 + . . . + cn x−m , ci ∈ m, and among all such relations we may choose one with minimal m. Without loss of generality m ≤ n: otherwise interchange x and x−1 . Then multiplying (23) by bn xn and subtracting from (22) gives another relation of the form (22) but with exponent smaller than n, contradiction. A subring R of a field K is a maximal subring if R intermediate between R and K . K and there is no ring Exercise: a) Show that any field K admits at least one maximal subring. b) Show that if K is not finite of prime order, then all maximal subrings are nonzero. c) Excluding the case in which K is finite of prime order, show that any maximal subring of K is a valuation ring of K (possibly a field). Exercise: Let K be a field, and R a valuation ring of K . Show TFAE: (i) R is a maximal subring of K . (ii) R has rank at most one. Theorem 408. Let K be a field and R ⊂ K a subring. Then the integral closure R of R in K is equal to the intersection of all valuation rings of K containing R. Proof. Let R be the intersection of all valuation rings of K containing R. Since each such ring is integrally closed in K and the intersection of a family of rings each integrally closed in K is again integrally closed in K , R is integrally closed in K , whence R ⊂ R. Conversely, let x ∈ K \ R. It suffices to find a valuation ring of K containing R but not x. Let y = x−1 . The ideal yR[y ] of R[y ] is proper: for if 1 = a1 y +. . .+an y n with ai ∈ R, then x would be integral over R. Let p be a maximal ideal of R containing y . By Lemma 407, there exists a valuation ring T of K such that R[y ] ⊂ K and mT ∩ R[y ] = p. Then y = x−1 ∈ mT , so x ∈ T . / 17.4. Another proof of Zariski’s Lemma. The following result is a close relative of Lemma 407. Lemma 409. Let K be a field and Ω an algebraically closed field. Let S be the set of all pairs (A, f ) with A is a subring of K and f : A → Ω, partially ordered by (A, f ) ≤ (A′ , f ′ ) ⇐⇒ A ⊂ A′ and f ′ |A = f. Then S contains maximal elements, and for any maximal element (B, g ), B is a valuation ring of K . Proof. An easy Zorn’s Lemma argument shows that S has maximal elements. Let (B, g ) be a maximal element. Put p = Ker(g ); since g (B ) is a subring of the field Ω, it is a domain and thus p is a prime ideal of B . By functoriality of localization, G extends to a homomorphism Bp → Ω. By maximality of (B, g ) we have Bp = B , COMMUTATIVE ALGEBRA 233 so that B is a local ring with maximal ideal p. If there existed an element x ∈ K which is transcendental over the fraction field of B , then B [x] is a polynomial ring and certainly g extends to B [x]. So K is algebraic over the fraction field of B . Next let x ∈ K × . We claim that either the ideal pB [x] or pB [x−1 ] is proper. Indeed this is proved exactly as in Lemma 407 above. Finally, we show that B is a valuation ring of K . Let x ∈ K • . Without loss of generality, we may assume that pB [x] is a proper ideal of B (otherwise replace x by x−1 ). Put B ′ = B [x]. By assumption, pB [x] is contained in a maximal ideal m of B ′ and m ∩ B = p. Hence the embedding of domains B → B ′ induces an embedding of fields k := B/p → B ′ /m = k ′ . Moreover k ′ is generated over k by the image of the algebraic element x, so k ′ /k is a finite degree field extension. So g induces an embedding k → Ω, and since Ω is algebraically closed, this extends to an embedding k ′ = B ′ /m → Ω. By maximality of B , this implies x ∈ B . Remark: It should be possible to consolidate Lemmas 407 and 409 into a single result. Let me know if you see how to do it. Proposition 410. Let A ⊂ B be domains with B finitely generated as an Aalgebra. Let β ∈ B • . There exists α ∈ A• satisfying the following property: any homomorphism f of A into an algebraically closed field Ω with f (α) ̸= 0 extends to a homomorphism f : B → Ω with f (β ) ̸= 0. Proof. Step 0: An easy induction argument on the number of generators reduces us to the monogenic case: B = A[x]. Step 1: Suppose that x is transcendental over A, i.e., B is a univariate polynomial ring over A. Write β = an xn + . . . + a1 x + a0 , ai ∈ A and put α = a0 . If f : A → Ω is such that f (α) ̸= 0, then since Ω is infinite, there exists ζ ∈ Ω such that f (an )ζ n + . . . + f (a1 )ζ + f (a0 ) ̸= 0. Using the universal polynomial of polynomial rings, we may uniquely extend f to a homomorphism from B to Ω by putting f (x) = ζ , and then f (β ) ̸= 0. Step 2: Suppose that x is algebraic over the fraction field of A. Then so is β −1 . Hence we have equations of the form an xm + . . . + a1 x + a0 , ai ∈ A a′ β −m + . . . + a′ β −1 + a′ , a′ ∈ A. m 1 0i Put α = an a′ . Suppose f : A → Ω is any homomorphism with f (α) ̸= 0. We may m extend f to a homomorphism from A[α−1 ] → Ω by mapping α−1 to f (α)−1 and then, by Lemma 409, to a homomorphism f : C → Ω for some valuation ring C containing A[α−1 ]. By construction x is integral over A[α−1 ]. Since C is integrally closed, x ∈ C . Thus C contains B and in particular β ∈ C . Similarly, β −1 is integral over A[α−1 ] so β −1 ∈ C . Thus β ∈ C × , so f (β ) ̸= 0. Restricting to B gives the desired homomorphism. Proof of Zariski’s Lemma: Let k be a field and B a field which is finitely generated as a k -algebra. We want to show that B is a finite field extension of B . Equivalently, it is enough to show that B/k is algebraic. In Proposition 410 take A = k , β = 1 and Ω to be an algebraic closure of k . 17.5. Discrete valuation rings. 234 PETE L. CLARK 17.5.1. Introducing DVRs. Proposition 411. For a valuation ring R with value group G, TFAE: (i) R is a PID. (ii) R is Noetherian. (iii) R is an ACCP-domain. (iv) G≥0 = {x ∈ G | x ≥ 0} is well-ordered. (v) G is isomorphic to (Z, ≤). Proof. (i) ⇐⇒ (ii) ⇐⇒ (iii) is a special case of Theorem 396. (iii) ⇐⇒ (iv): a totally ordered set is well-ordered iff there are no infinite strictly descending chains. But an infinite strictly descending chain in the value group of R gives rise to an infinite strictly ascending chain of principal ideals in R, and conversely. (iv) =⇒ (v): First suppose G is Archimedean, so G → R: this endows G with a topology. If G is discrete, it is generated by its least positive element hence is order-isomorphic to Z. If G is not discrete, there exists an infinite strictly decreasing sequence of positive elements of G converging to 0, so G≥0 is not well-ordered. Next suppose that G is not Archimedean, and choose x, y > 0 such that for all n ∈ Z+ , nx < y . Then {y − nx}n∈Z+ is an infinite strictly descending sequence in G≥0 , contradicting well-ordering. (v) =⇒ (iv): famously, the standard ordering on Z≥0 is a well-ordering. Exercise X.X: Show directly that a local PID is a valuation ring with value group Z. A valuation ring satisfying the equivalent conditions of Proposition 411 is called a discrete valuation ring (or, sometimes, a DVR). 17.5.2. Further characterizations of DVRs. In many ways, discrete valuation rings are – excepting only fields – the simplest class of rings. Nevertheless they have an important role to play in algebra and arithmetic and algebraic geometry. One reason for this is as follows: every DVR is a one-dimensional Noetherian local ring. The converse does not hold. Example: Let k be a field, and let R be the k -subalgebra of k [t] generated by t2 and t3 . This is a one-dimensional Noetherian domain; the ideal m generated by t2 and t3 is a nonprincipal maximal ideal. Indeed, even in the localization Rm the ideal mRm is not principal: consider what its order at 0 would be with respect to the valuation ordt on k (t): it would have to be 1, but there is no such element in Rm . The question then is to find necessary and sufficient conditions for a one-dimensional Noetherian local domain to be a DVR. As we have seen, being a PID is enough, but again, this is not very useful as whether a one-dimensional domain is a PID is difficult to check in practice. Remarkably, it turns out if a local, one-dimensional Noetherian domain has any one of a large number of good properties, it will necessarily be a DVR. Here is the theorem. Theorem 412. (Recognition Theorem for DVRs) Let R be a one-dimensional Noetherian local domain, with maximal ideal m. TFAE: m (i) R is regular: the dimension of m2 as an R/m-vector space is 1. COMMUTATIVE ALGEBRA 235 (ii) m is principal. (iii) R is a PID. (iv) R is a UFD. (v) R is integrally closed. (vi) Every nonzero ideal is of the form mn for some n ∈ N. Proof. (i) ⇐⇒ (ii): Choose t ∈ π \ π 2 . By assumption, t generates m/m2 , so by Nakayama’s Lemma t generates m. Conversely, if m is monogenic as an R-module, certainly m/m2 is monogenic as an R/m-module. Evidently (iii) =⇒ (ii). Proposition 385 gives (ii) =⇒ (iii) and also (ii) =⇒ (vi). Moreover (iii) ⇐⇒ (iv) by Proposition 380 and (iv) =⇒ (v) by 354. Next, for all n ∈ N we have (π )n /(π )n+1 ∼ R/m, thus R is regular. = (vi) =⇒ (i): Assume that dimR/m m/m2 > 1. Choose u ∈ m \ m2 . Then we have m ⟨u, m2 ⟩ m2 . So we have (i) ⇐⇒ (ii) ⇐⇒ (iii) ⇐⇒ (iv) ⇐⇒ (vi) =⇒ (v). Finally, we show (v) =⇒ (ii): Let 0 ̸= x ∈ m. Since m is the only prime ideal containing (x) we must have r((x)) = m. Since R/(x) is Noetherian, its radical, m/(x), is nilpotent, so there is a unique least n ∈ Z+ such that mn ⊂ (x). Let y ∈ mn−1 \ (x) and consider the element q = x of the fraction field K of R. Since y y y ∈ (x), q −1 = x ∈ R; since R is assumed integrally closed in K , q −1 is not integral / / over R. Then we must have q−1 m is not contained in m, for otherwise m would be a faithful R[q −1 ]-module which is finitely generated as an R-module, contradicting y Theorem 310. But by construction, q −1 m = x m ⊂ R, hence q −1 m = R and then m = Rx = (x). 17.5.3. Modules over DVRs. Lemma 413. Let R be a DVR with uniformizing element π , and let a ∈ Z+ . Then the ring Ra = R/(π a ) is self-injective – i.e., Ra is an injective Ra -module. Exercise X.X: Prove Lemma 413. (Hint: Baer’s Criterion!) Theorem 414. Let R be a DVR with uniformizing element π , and let M be a nonzero finitely generated R-module. a) There is N ∈ N and positive integers n, a1 ≥ a2 ≥ . . . ≥ an such that (24) M ∼ RN ⊕ = n ⊕ R/(π ai ). i=1 b) The numbers N, n, a1 , . . . , an are invariants of the isomorphism class of the module M : i.e., they are the same for any two decompositions of M as in (24) above. Proof. Step 0: Consider the canonical short exact sequence 0 → M [tors] → M → M/M [tors] → 0. Since M is a finitely generated module over a Noetherian ring, M [tors] is finitely generated. Moreover, M/M [tors] is a finitely generated torsionfree module over a PID, hence is free (Proposition 66). Moreover, we know that the rank of a free module over any (commutative!) ring is well-defined (when R is a domain with fraction field K , the proof is especially easy: the rank of a free module M is 236 PETE L. CLARK dimK M ⊗R K ), so the invariant N in the statement of the theorem is precisely the rank of M/M [tors]. Moreover, since M/M [tors] is free – hence projective – the sequence splits, so M = RN ⊕ M [tors]. We are reduced to the case of a nonzero finitely generated torsion module M . Step 1: The annihilator of M is an ideal of R, of which there aren’t so many: it must be (π a1 ) for some a1 ∈ Z+ . Thus M may be viewed as a faithful Ra1 = R/(π a1 )module. Moreover, choosing an element x of M which is not annihilated by π a1 −1 , the unique Ra1 -module map Ra1 → M which sends 1 to m is an injection. Taking M ′ = M/Ra1 , we get a short exact sequence 0 → Ra1 → M → M ′ → 0. By Lemma 413, Ra1 is an injective Ra1 -module, so the sequence splits: M ∼ Ra ⊕ M ′ . = 1 Step 2: Since M is finitely generated over Ra1 , it is a quotient of some Artinian M Ra1 -module Ra1 , hence by Theorem 188 M is Artinian. Moreover M is a finitely generated module over the Noetherian ring, so M is also Noetherian. By Theorem 198, this means that M has finite length as an R-module. Hence so does its direct summand M ′ and indeed clearly the length of M ′ is less than the length of M . This completes the proof of part a) by induction. Step 3: So far we have that a finitely generated torsion R-module is of the form ⊕n ai a1 i=1 R/(π ) with positive integers a1 ≥ a2 ≥ . . . ≥ an , and with ann(M ) = (π ). In order to prove the uniqueness statement of part b), it suffices to prove that for all 0 < b ≤ a, R/(π b ) is an indecomposable R/(π a )-module. If so, then M∼ = n ⊕ R/(π ai ) i=1 is simply the decomposition of the finite length module M into indecomposables described in the Krull-Schmidt Theorem: in particular, since clearly R/(π a ) ∼ = R/(π b ) implies a = b (consider annihilators), it is unique up to permutation of the factors. So suppose that R/(π a ) = M1 ⊕ M2 with M1 , M2 nonzero. If π a does not annihilate M1 , then as above we can find a split embedding R/(π a ) → M1 , which contradicts the fact that the length of M1 must be smaller than the length of R/(π a ). So M1 – and similarly M2 – is annihilated by π a−1 and thus R/(π a ) would be annihilated by π a−1 , a contradiction. Remark: Theorem 414 is nothing else than the fundamental structure theorem for modules over a PID in the special case in which the PID has a unique maximal ideal. But we have not given a proof of this structure theorem for PIDs in these notes, whereas later we will want to use it to prove a more general structure theorem for torsion modules over a Dedekind domain. However, in both cases it is easy to reduce to the local situation, and thus we will get an independent proof. 18. Normalization theorems We work in the following situation: R is an integrally closed domain with fraction field K , L/K is a field extension, and S = IL (R) is the integral closure of R in L. In more geometric language, S is the normalization of R in the extension L/K . As above, we may as well assume that L/K is algebraic, since in the general COMMUTATIVE ALGEBRA 237 case, if we let L′ = IK (L) be the algebraic closure of K in L, then S is contained in L′ anyway. So let us assume this. Then we know that S is integrally closed with fraction field L. We also know that the Krull dimensions of S and R coincide. The major questions are the following: (Q1) Is S finitely generated as an R-module? (Q2) Is S Noetherian? (Q3) If not, then can anything nice be said about S ? Note that if R is Noetherian, then an affirmative solution to (Q1) implies an affirmative answer to (Q2). (If R is not Noetherian there is no reason to think that S should be – e.g. take L = K ! – so we are “only” interested in (Q2) in case R is Noetherian.) Also, the example of R = Z, K = Q, L = Q above shows that both (Q1) and (Q2) may have a negative answer if [L : K ] is infinite. So we are most interested in the finite case, although, as we shall see, there is something to say in the infinite case as well. 18.1. The First Normalization Theorem. The first, and easiest, result is the following: Theorem 415. (First Normalization Theorem) Let R be an integrally closed domain with fraction field K , L/K a finite separable field extension, and S = IL (R). a) There exists a K -basis x1 , . . . , xn of L such that S is contained in the Rsubmodule generated by x1 , . . . , xn . b) Therefore if R is Noetherian, S is finite as an R-module. c) If R is a PID, then S is a free R-module of rank [L : K ]. Proof. By the proof of Proposition 319, for any x ∈ L, there exists 0 ̸= r ∈ R such that rx ∈ S . Therefore there exists a K -basis u1 , . . . , un of L such that ui ∈ S ∑ for all i.59 Now take x ∈ S and write x = i bi ui with bi ∈ K . Since L/K is separable there are n = [L : K ] distinct K -embeddings of L into K , say σ1 , . . . , σn , and the discriminant ∆ = ∆(u1 , . . . , un ) = (det(σj (ui )))2 is nonzero. We may put √ ∆) = det(σj (ui )). For all 1 ≤ j ≤ n we have ∑ σj (x) = bi σj (ui ). i Using Cramer’s rule, we may solve for the bi to get ∑ ∑√ √ ∆bi = dij σj (x), dbi = ddij σj (x), j j where the dij ’s are certain polynomials in the σj (ui ) with integer coefficients. This √ shows that ∆bi and ∆bi are integral over R. Since ∆ ∈ K and R is integrally closed, we have ∆bi ∈ A. Therefore S is contained in the R-span ⟨ u1 , . . . , un ⟩R , ∆ ∆ establishing part a). Since over a Noetherian ring, submodules of finitely genereated modules are themselves finitely generated, this implies b). Our work so far shows that S is a submodule of a free rank n R-module, so over an integral domain this 59Note that we have not yet used the separability hypothesis, so this much is true in the case of an arbitrary finite extension. 238 PETE L. CLARK implies that S is itself a free R-module of rank at most n. Since S ⊗R K = L, the rank must be exactly n. This has the following important result, which is the first of three fundamental finiteness theorems in algebraic number theory, the existence of a finite integral basis for the ring of integers of any algebraic number field: Corollary 416. Let R = Z, K = Q, L a number field of degree n. Then the ring ZL = Z ∩ K of all algebraic integers lying in L, is an integrally closed, Noetherian domain of Krull dimension one which is, as a Z-module, free of rank n. Proof. The ring ZL is nothing else than IL (Z), so by Proposition 320, it is integrally closed in its field of fractions L. Since Z is a PID and L/Q is finite separable, Theorem 415 applies to show that ZL ∼ Zn as a Z-module. Being a finitely generated = Z-module, still more is it a finitely generated algebra over the Noetherian ring Z, so it is itself Noetherian. Since Z, like any PID, has Krull dimension one and ZL is an integral extension of Z, by Corollary 326 ZL also has Krull dimension one. A Dedekind domain is an integral domain R which is Noetherian, integrally closed and of Krull dimension at most one. We will explore this concept in loving detail later on, but notice for now that Corollary 416 shows in particular that the ring ZL of integers of any algebraic number field L is a Dedekind domain. In fact, the argument establishes that the normalization S of any Dedekind domain R in a finite separable field extension L/K is again a Dedekind domain which is finitely generated as an R-module. What about the nonseparable case? Give Kaplansky’s example. COMPLETE ME! ♣ 18.2. The Second Normalization Theorem. Theorem 417. (Second Normalization Theorem) Let R be an integral domain with fraction field K . Suppose that at least one of the following holds: • R is absolutely finitely generated i.e., finitely generated as a Z-algebra – or • R contains a field k and is finitely generated as a k -algebra. Let L/K a finite field extension. Then S = IL (R) is a finitely generated R-module. Proof. First spupose that R is a finitely generated algebra over a field k . Step 0: We may assume that L/K is normal. Indeed, let M be the normal closure of L/K , so M/K is a finite normal extension. Let T be the integral closure of R in M . If we can show that T is finitely generated over R, then, since R is Noetherian, the finitely submodule S is also finitely generated over R. Step 1: We will make use of the field-theoretic fact that if M/K is normal and L is the maximal purely inseparable subextension of M/K , then M/L is separable [FT, §6.3]. Let S be the integral closure of R in L and T the integral closure of R in T . Then T is module-finite over R iff T is module-finite over S and S is module-finite over R. Suppose we can show that S is module-finite over R. Then S is a finitely generated R-algebra so S is a Noetherian integreally closed domain, and the module finiteness of T over S follows from Theorem 415. Thus we are redueced to the case in which L/K is purely inseparable, say [L : K ] = q = pa . Step 2: By Noether Normalization, R is module-finite over a polynomial ring COMMUTATIVE ALGEBRA 239 k [t1 , . . . , td ]. If S is module-finite over k [t1 , . . . , td ], then certainly it is modulefinite over the larger ring R. Thus we may assume without loss of generality that R = k [t1 , . . . , td ], K = k (t1 , . . . , td ). In particular we may assume that R is integrally closed (in K ). For all a ∈ L, NL/K (a) = aq ∈ K . Let k ′ /k be the extension obtained by adjoining the q th roots of the coefficients of the minimal polynomials of 1/q 1/q a finite set of generators of L/K , so k ′ /k is finite, so L ⊂ k ′ (t1 , . . . , td ). So it is 1/q 1/q enough to show that the integral closure of k [t1 , . . . , td ] in k ′ (t1 , . . . , td ) is finite over k [t1 , . . . , td ]. But in this case the integral closure can be computed exactly: it 1/q 1/q is k ′ [t1 , . . . , td ] (indeed it is at least this large, and this ring is a UFD, hence integrally closed), which is finite over k [t1 , . . . , td ]. 18.3. The Krull-Akizuki Theorem. COMPLETE ME! ♣ 19. Fractional ideals 19.1. Definition and first properties. Let R be an integral domain with fraction field K . A fractional ideal of R is a nonzero R-submodule I of K for which there exists 0 ̸= a ∈ R such that aI ⊂ R 1 – or equivalently, if I ⊂ a R. When one is talking about fractional R-ideals, one inevitably wants to compare them to ideals of R in the usual sense, and for this it is convenient to speak of an integral R-ideal, i.e., an R-submodule of R. Exercise: Show that any finitely generated R-submodule of K is a fractional ideal. Comment: Some references define a fractional R-ideal to be a finitely generated R-submodule of K , but this seems wrong because we certainly want every nonzero integral ideal of R to be a fractional ideal, but if R is not Noetherian then not every integral ideal will be finitely generated. (It is not such a big deal because most of these references are interested only in invertible fractional ideals – to be studied shortly – and one of the first things we will see is that an invertible fractional ideal is necessarily finitely generated as an R-module.) We denote the set of all fractional ideals of R by Frac(R). Let I, J ∈ Frac(R). Theorem 418. Let I, J, K be fractional ideals in an integral domain R. a) Then I ∩ J = {x ∈ K | x ∈ I and x ∈ J }, I + J = {x + y | x ∈ I, y ∈ J, n ∑ IJ = { xi yi , | xi ∈ I, yi ∈ J }, i=1 (I : J ) = {x ∈ K | xJ ⊂ I } are all fractional ideals. b) We may partially order Frac(R) under inclusion. Then the greatest lower bound of I and J is I ∩ J and the least upper bound of I and J is I + J . 240 PETE L. CLARK c) If I ⊂ J , then IK ⊂ JK . d) R itself is a fractional ideal, and R · I = R. Thus the fractional ideals form a commutative monoid under multiplication. Proof. a) It is immediately verified that I ∩ J , I + J , IJ and I : J are all Rsubmodules of K . It remains to be seen that they are nonzero and can be scaled 1 to lie inside R. Suppose I ⊂ a R and J ⊂ 1 R. Then: b 1 0 I ⊂ I + J ⊂ ab R, so I + J is a fractional ideal. 1 0 I J ⊂ I ∩ J ⊂ ab R, so IJ and I ∩ J are fractional ideals. Since I ∩ R is a fractional ideal, there exists a nonzero c ∈ R lying in I . Then for 1 y ∈ J , c y ∈ cR ⊂ I , so c ∈ (I : J ). Similarly, if 0 ̸= d ∈ J , then ad I : J ⊂ R. Parts b b b), c) and d) can be easily verified by the reader. Proposition 419. All the above operations on fractional ideals commute with localization: that is, if S ⊂ R• is a multiplicatively closed subset, then (S −1 (I ∩ J ) = S −1 I ∩ S −I J, S −1 (I + J ) = S −1 I + S −1 J, S −1 (IJ ) = (S −1 I )(S −1 J ), S −1 (I : J ) = (S −1 I : S −1 J ). Exercise: Prove Proposition 419. 19.2. Principal Fractional Ideals. A fractional ideal is said to be principal if it is of the form xR for some x = a b ∈ K •. The following fact is easy but extremely important. Proposition 420. For a fractional ideal I of R, TFAE: (i) I is principal. (ii) I is monogenic as an R-module. (iii) I ∼R R. = Proof. By definition a monogenic R-module M is one of the form Rx for some x ∈ M , so the equivalence of (i) and (ii) is immediate. Certainly R is monogenic as an R-module. Conversely, suppose I = Rx for x ∈ K × . Then multiplication by x−1 gives an isomorphism to R. (Another way to look at it is that a module M over a domain R is isomorphic to R itself iff it is monogenic and torsionfree, and a principal fractional ideal has both of these properties.) If xR is a principal fractional ideal, so is x−1 R, and we have (xR)(x−1 R) = R. Thus, in the monoid Frac(R), every principal fractional ideal xR is a unit, with inverse x−1 R. Let Prin(R) denote the set of all principal fractional ideals of the domain R. Exercise: Show that Prin(R) is a subgroup of Frac(R), and we have a short exact sequence 1 → R× → K × → Prin(R) → 1. COMMUTATIVE ALGEBRA 241 Exercise: Define the ideal class monoid C (R) = Frac(R)/ Prin(R). a) Show that C (R) is well-defined as a commutative monoid. b) Show that √(R) is trivial iff R is a PID. C c) Let R = Z[ −3]. Show that C (R) is a finite commutative monoid which is not a group. It is somewhat disappointing that for a general domain, C (R) need only be a commutative monoid. In the next section we “repair” this by defining the Picard group Pic(R). 19.3. Invertible Fractional Ideals and the Picard Group. Like any monoid, Frac(R) has a group of units, i.e., the subset of invertible elements. Explicitly, a fractional ideal I is invertible if there exists another fractional ideal J such that IJ = R. We denote the group Frac(R)× of invertible fractional ideals by Inv(R). Exercise: Let I1 , . . . , In be fractional ideals of R. Show that the product I1 · · · In is invertible iff each Ii is invertible. (Note: this has nothing to do with fractional ideals, but is rather a fact about the units in a commutative monoid.) It turns out that to every fractional ideal we can attach another fractional ideal I ∗ which will be the inverse of I iff I is invertible. Namely, for I ∈ Frac(R), put I ∗ = (R : I ) = {x ∈ K | xI ⊂ R}. Immediately from the definition we get II ∗ ⊂ R. Lemma 421. For a fractional ideal I , TFAE: (i) I is invertible. (ii) II ∗ = R. Proof. (i) =⇒ (ii): As above, for any fractional ideal I we have II ∗ ⊂ R. Now suppose there exists some fractional ideal J such that IJ = R, then J ⊂ (R : I ) = I ∗ , so R = IJ ⊂ II ∗ . (ii) =⇒ (i) is obvious. Proposition 422. Let I be an invertible fractional ideal. Then as an R-module I is a rank one projective module. Proof. Step 1: We show that an invertible fractional ideal I is a finitely generated ∑n projective module. Since II ∗ = R, we may write 1 = i=1 xi yi with xi ∈ I and yi ∈ I ∗ . For 1 ≤ i ≤ n, define fi ∈ Hom(I, R) be fi (x) = xyi . Then for all x ∈ I , ∑ ∑ x= xxi yi = xi fi (x). i i By the Dual Basis Lemma, I is a projective R-module generated by x1 , . . . , xn . Step 2: Recall that to show that I has rank one, we must show that for all p ∈ Spec R, Ip is free of rank one over Rp . But since projective implies locally free, we 242 PETE L. CLARK know that Ip is a free Rp -module of some rank, and it is quite elementary to show that for any ring R and any ideal I , I cannot be free of rank greater than one over R. Indeed, if so I would have two R-linearly independent elements x and y , which is absurd, since yx + (−x)y = 0. Conversely: Proposition 423. Let I be a nonzero fractional ideal of R which is, as an Rmodule, projective. Then I is invertible. Proof. We have the inclusion ι : II ∗ ⊂ R which we wish to show is an equality. This can be checked locally: i.e., it is enough to show that for all p ∈ Spec R, ∗ ιp : Ip Ip → Rp is an isomorphism. By Proposition 419, it is equivalent to show that Ip (Ip )∗ → Rp is an isomorphism, but since I is projective, by Kaplansky’s Theorem Ip is free. As above, being a nonzero ideal, it is then necessarily free of rank one, i.e., a principal fractional ideal xRp . It follows immediately that (Ip )∗ = x−1 Rp and thus that the map is an isomorphism. To sum up: Theorem 424. Let I be a nonzero fractional ideal for a domain R. Then I is invertible iff it is projective, in which case it is necessarily projective of rank one. For any R-module M , the R-dual is defined to be M ∨ = Hom(M, R). There is a canonical R-bilinear map T : M ∨ × M → R obtained by mapping (f, x) → f (x). This induces an R-linear map T : M ∨ ⊗R M → R. Let us say that an R-module M is invertible if D is an isomorphism. Proposition 425. Consider the following conditions on an R-module M . (i) M is rank one projective. (ii) M is invertible. (iii) There exists an R-module N and an isomorphism T : M ⊗R N ∼ R. = Then (i) =⇒ (ii) =⇒ (iii) always, and (iii) =⇒ (i) if M is finitely generated. Proof. (i) =⇒ (ii): As usual, we have a globally defined map T : M ∨ ⊗ M → R so that it suffices to check locally that is an isomorphism, but M is locally free so this is easy. (ii) =⇒ (iii) is immediate. (iii) =⇒ (i): Since M is finitely generated, by Theorem 308 to show that M is projective it is enough to show that for all p ∈ Spec R Mp is free of rank one. Thus we may as well assume that (R, m) is a local ring with residue field R/m = k . The base change of the isomorphism T to R/m is an isomorphism (recall that tensor product commutes with base change) Tk : M/mM ⊗k N/mN → k. This shows that dimk M/mM = dimk N/mN = 1, so in particular M/mM is monogenic as an R/m-module. By Nakayama’s Lemma the lift to R of any generator x of M/mM is a generator of M , so M is a monogenic R-module and is thus isomorphic to R/I for some ideal I . But indeed I = ann(M ) ⊂ ann(M ⊗R N ) = ann(R) = 0, so M ∼ R/(0) = R is free of rank one. = Theorem 426. Let I and J be invertible fractional ideals. Then there is a canonical isomorphism of R-modules ∼ I ⊗R J → IJ. COMMUTATIVE ALGEBRA 243 Proof. The natural multiplication map I × J → IJ is R-bilinear so factors through an R-module map m : I ⊗R J → IJ . Again, once we have a globally defined map, to see that it is an isomorphism it is enough to check it locally, i.e., that for all p ∈ Spec R, ∼ mp : Ip ⊗Rp Jp → Ip Jp and we are thus allowed to assume that I and J are principal fractional ideals. This makes things very easy, and we leave the endgame to the reader. Corollary 427. Let I and J be invertible fractional R-ideals. TFAE: (i) There exists x ∈ K × such that xI = J . (ii) I ∼R J , i.e., I and J are congruent as R-modules. = Proof. (i) =⇒ (ii): If J = xI , then multiplication by x gives an R-module isomorphism from I to J . (ii) =⇒ (i): Since I ∼R J we have = I −1 J ∼ I −1 ⊗R J ∼ I −1 ⊗R I ∼ II −1 = R. = = = By Proposition 420, I −1 J is a principal fractional ideal, i.e., there exists x ∈ K × such that I −1 J = xR. Multiplying through by I , we get xI = J . Proposition 428. Let M be a rank one projective module over a domain R. Then there exists a fractional R-ideal I such that M ∼R I . = Proof. Let K be the fraction field of R. Since M is projective, it is flat, and so tensoring the injection R → K with M we get an injection f : M = R ⊗R M → M ⊗R K ∼ K , the last isomorphism since M is locally free of rank 1. Thus = ∼ f : M → f (M ), and f (M ) is a finitely generated R- submodule of K and thus a fractional R-ideal. Putting together all the pieces we get the following important result. Theorem 429. Let R be a domain. The following two commutative groups are canonically isomorphic: (i) Inv(R)/ Prin(R) with [I ][J ] := [IJ ]. (ii) Isomorphism classes of rank one projective R-modules under tensor product. We may therefore define the Picard group Pic R to be either the group of invertible fractional ideals modulo principal fractional ideals under multiplication or the group of isomorphism classes of rank one projective modules under tensor product. 19.4. More on invertible ideals. Lemma 430. In any domain R, let P1 , . . . , Pk be a set of invertible prime ideals and let Q1 , . . . , Ql be any set of prime ideals. Suppose that k ∏ i=1 Pi = l ∏ Qj . j =1 Then i = j and there exists some permutation σ of the set {1, . . . , k } such that for all 1 ≤ i ≤ k we have Pi = Qσ(i) . In other words, prime factorization is unique for products of invertible primes. 244 PETE L. CLARK Proof. Assume without loss of generality that P1 does not strictly contain any ∏ Pi . Since j Qj ⊂ P1 , some Qj , say Q1 , is contained in P1 . Similarly, since ∏ i Pi ⊂ Q1 , there exists i such that Pi ⊂ Q1 . Thus Pi ⊂ Q1 ⊂ P1 . By our assumption on the minimality of P1 , we have P1 = Pi = Q1 . We can thus cancel P1 = Q1 and obtain the result by induction. Lemma 431. Let R be an integrally closed Noetherian domain with fraction field K , and let I be a fractional R-ideal. Then [I : I ] := {x ∈ F | xI ⊂ I } = R. Proof. Clearly R ⊂ [I : I ]. Conversely, let x ∈ [I : I ]. Then I is a faithful R[x]-module which is finitely generated over R, so x is integral over R. Lemma 432. Let R be a domain with fraction field K , S ⊂ R \{0} a multiplicative subset, and I, J fractional R-ideals. a) We have (I + J )S = IS + JS . b) (IJ )S = IS JS . c) (I ∩ J )S = IS ∩ JS . d) If I is finitely generated, then (I ∗ )S = (IS )∗ . Proof. Parts a) and b) are immediate and are just recorded for future reference. For part c), we evidently have (I ∩ J )S ⊂ IS ∩ JS . Conversely, let x ∈ IS ∩ JS , so x = si1 = sj2 with i ∈ I , j ∈ J , s1 , s2 ∈ S . Put b = a1 s2 = a2 s1 ∈ I ∩ J ; then x = s1bs2 ∈ (I ∩ J )S , establishing part c). For part d), note first that (I + J )∗ = I ∗ ∩ J ∗ . Also if 0 ̸= x ∈ K , then (Rx)S = RS x. Hence if I = Rx1 + . . . + Rxn , then ∩n 1 ∩n 1 IS = RS x1 + . . . + RS xn , so (IS )∗ = i=1 xi RS . On the other hand, I ∗ = i=1 xi R, and thus part c) we have n ∩1 ∗ (I )S = RS = (IS )∗ . x i=1 i Lemma 433. Let R be a Noetherian domain of Krull dimension at most one. Let I be a proper, nonzero ideal of R. Then (R : I ) R. Proof. COMPLETE ME! ♣ The following result gives information about when a prime ideal is invertible. Proposition 434. Let R be a Noetherian domain, and p a nonzero prime ideal of R. If p is invertible, then it has height one and Rp is a DVR. Proof. Since p is invertible, Rp is a Noetherian local domain with a principal maximal ideal pRp . By Theorem 412, Rp is a DVR, and thus p has height one. 19.5. Divisorial ideals. Let R be a domain with fraction field K . Let Frac(R) be the monoid of nonzero fractional R-ideals in K , under multiplication. For I, J ∈ Frac(R), we write I < J if every principal fractional ideal Ra which is contained in I is also contained in J . This gives a preordering on Frac(R). Let R be the associated equivalence relation, i.e., I ∼ J if I ≤ J and J ≤ I . We write D(R) = Frac(R)/ ∼. Elements of D(R) are called divisors on R. For a fractional ideal I , we denote its image in D(R) by div(I ), and for a principal fractional ideal aR, we write simply div(a). Such elements are called principal divisors. COMMUTATIVE ALGEBRA 245 Let I ∈ Frac(R). By hypothesis there exists 0 ̸= d ∈ R such that I ⊂ d−1 R. From this it follows that the intersection of all principal fractional ideals containing I is itself a fractional ideal. We define this intersection to be ∩ ˜ d−1 R. I= d∈R• | I ⊂d−1 R ˜ ˜ ˜ Evidently we have I = I , so I → I may be viewed as some sort of projection operator on fractional ideals. Indeed we say that a fractional ideal I is divisorial if ˜ ˜ I = I . It is easy to see that I is indeed the least divisorial fractional ideal contain˜ ing I . Moreover, it is the unique divisorial fractional ideal such that div(I ) = div(I ). Exercise: If I is a divisorial fractional ideal and x ∈ K • , then Ix is a divisorial fractional ideal. Exercise X.X: Let {Ii } be a family of divisorial fractional ideals such that I = is nonzero. Then I is a divisorial fractional ideal. ∩ i Ii Lemma 435. Let I, J in Frac(A). a) We have ∩ J :I= x−1 J. x∈I • b) If J is divisorial, then so is (J : I ). Proof. a) Let y ∈ K . Then y ∈ J : I iff yI ⊂ J iff for all x ∈ I • , yx ∈ J iff for all x ∈ I • , y ∈ x−1 J . b) Since J is divisorial, by Exercise X.X so is each x−1 J . Thus J : I is an intersection of divisorial fractional ideals which is, by Theorem 418, nonzero, hence by Exercise X.X J : I is a divisorial fractional ideal. Proposition 436. Let I, J ∈ Frac(R). a) We have div(I ) = div(J ) iff R : I = R : J . ˜ b) I = R : (R : I ). Proof. a) Let P (I ) be the set of principal fractional ideals containing I . Thus Rx ∈ P (I ) ⇐⇒ x−1 I ⊂ R ⇐⇒ x−1 ∈ R : I . Thus div(I ) = div(J ) ⇐⇒ P (I ) = P (J ) ⇐⇒ R : I = R : J . b) Since I (R : I ) ⊂ R, I ⊂ (R : (R : I )). Substituting (R : I ) for I in these relations, we get (R : I ) ⊂ R : (R : (R : I )). On the other hand, from I ⊂ R : (R : I ) we get (R : I ) ⊃ R : (R : (R : I )). Thus R : I = R : (R : (R : I ))), thus using part a) we get div(I ) = div(R : (R : I )). However, by the previous Lemma R : (R : I ) is divisorial, so it is the unique divisorial ideal whose divisor is ˜ equal to div(I ), i.e., I = R : (R : I ). Exercise: Interpret the above results in terms of a Galois connections between Frac(R) and D(R). 246 PETE L. CLARK Proposition 437. (Bourbaki, p. 477) a) In D(R) every nonempty set which is bounded above admits a least upper bound. Explicitly, if (Ii ) is a nonempty family of fractional ideals which is bounded above, ∩˜ then supi (div Ii ) = div( i Ii ). b) In D(R) every nonempty set which is bounded below admits a greatest lower bound. Explicitly, if (Ji ) is a nonempty family of fractional ideals which is bounded ∑ below, then inf i (div Ji ) = div( i Ii ). c) D(R) is a lattice. Proof. Bourbaki, p. 477. COMPLETE ME! ♣ 19.6. The Divisor Class Group. 20. Dedekind domains A Dedekind domain is an integral domain which is Noetherian, integrally closed, and of dimension at most one. A Dedekind domain has dimension zero iff it is a field. Although we endeavor for complete precision here (why not?), the reader should be warned that in many treatments the zero-dimensional case is ignored, when convenient, in statements of results. 20.1. Characterization in terms of invertibility of ideals. Theorem 438. For an integral domain R, TFAE: (i) R is Dedekind: Noetherian, integrally closed of dimension at most one. (ii) Every fractional R-ideal is invertible. Proof. (i) =⇒ (ii): Let R be a Noetherian, integrally closed domain of dimension at most one, and let I be a fractional R-ideal. Then II ∗ ⊂ R and hence also II ∗ (II ∗ )∗ ⊂ R, so I ∗ (II )∗ ⊂ I ∗ . It follows from Lemma 431 that (II ∗ )∗ ⊂ R; moreover, since II ∗ ⊂ R, Lemma 433 implies II ∗ = R, i.e., I is invertible. (ii) =⇒ (i): Since invertible ideals are finitely generated, if every nonzero ideal is invertible, then R is Noetherian. Let p be a nonzero, nonmaximal prime ideal of R, so that there exists a maximal ideal m which 0 p m. By the mantra “to contain is to divide” for invertible fractional ideals, there exists some invertible integral ideal I such that p = mI . Suppose that I ⊂ p. Then I = RI ⊃ mI = p, so we would have p = I and then m = R, contradiction. Then there exists x ∈ m \ p and y ∈ I \ p such that xy ∈ p, contradicting the primality of p. b Finally, we check that R is integrally closed: let x = c be a nonzero element of the fraction field K of R which is integral over R, so there exist a0 , . . . , an−1 ∈ R such that xn + an−1 xn−1 + . . . + a1 x + a0 = 0. Let M be the R-submodule of K generated by 1, x, . . . , xn−1 ; since M is finitely generated, it is a fractional R-ideal. We have M 2 = M , and thus – since M is invertible – M = R. It follows that x ∈ R. Recall that a ring R is hereditary if every ideal of R is a projective R-module. Corollary 439. A domain R is hereditary iff it is a Dedekind domain. Proof. By Theorem 438 a domain R is a Dedekind domain iff every fractional ideal of R is invertible, and clearly the latter condition holds iff every nonzero integral ideal of R is invertible. Moreover, by Theorem 424, a nonzero ideal of a ring is invertible iff it is projective as an R-module. COMMUTATIVE ALGEBRA 247 20.2. Ideal factorization in Dedekind domains. Here we will show that in a Dedekind domain every nonzero integral ideal factors uniquely into a product of primes and derive consequences for the group of invertible ideals and the Picard group. (The fact that factorization – unique or otherwise! – into products of primes implies invertibility of all fractional ideals – is more delicate and will be pursued later.) Lemma 440. Let I be an ideal in a ring R. If there exist J1 , J2 ideals of R, each strictly containing I , such that I = J1 J2 , then I is not prime. Proof. Choose, for i = 1, 2, xi ∈ Ji \ I ; then x1 x2 ∈ I , so I is not prime. Lemma 441. A nonzero ideal in a Noetherian domain contains a product of nonzero prime ideals. Proof. Assume not, and let I be a nonzero ideal which is maximal with respect to the property of not containing a product of nonzero prime ideals. Then I is not prime: there exist x1 , x2 ∈ R \ I such that x1 x2 ∈ I . Now put, for i = 1, 2, Ii := ⟨I, xi ⟩, so that I Ii and I ⊃ I1 I2 . By maximality of I , I1 ⊃ p1 · · · pr and I2 ⊃ q1 · · · qs (with pi , qj prime for all i, j ), and then I ⊃ p1 · · · pr q1 ⊃ qs , contradiction. Theorem 442. Every proper integral ideal in a Dedekind domain has a unique factorization into a product of of prime ideals. Proof. After Lemma 430 it suffices to show that a nonzero proper integral ideal I in a Dedekind domain R factors into a product of primes. Suppose not, so the set of ideals which do not so factor is nonempty, and (as usual!) let I be a maximal element of this set. Then I is not prime, so in particular is not maximal: let p be a maximal ideal strictly containing I , so I = pJ . Then J = p−1 I strictly contains I so factors into a product of primes, hence I does. Notation: If I is any nonzero integral ideal of I and p is any nonzero prime ideal of a Dedekind domain R, then we may define non-negative integers ordp (I ) via the prime factorization ∏ I= pordp (I ) . p Note that the product extends formally over all primes, but as any given I is divisible by only finitely many prime ideals, all but finitely many exponents are zero, so it is really a finite product. Corollary 443. Let R be a Dedekind domain. a) The monoid M(R) of nonzero integral ideals is a free commutative monoid on the set of nonzero prime ideals. b) The fractional ideals form a free commutative group on the set of prime ideals: ⊕ Frac(R) = Z. 0̸=p ∈ Spec R Proof. Part a) is simply the statement of unique factorization into prime elements in any commutative monoid. In the group I (R) of all fractional ideals, the subgroup G generated by the nonzero primes is a free commutative group on the primes: this just asserts that for primes p1 , . . . , pr and integers n1 , . . . , nr , the equation 248 PETE L. CLARK pn1 · · · pnr = R implies n1 = . . . = nr = 0, which is easily seen – e.g. by localizing. r 1 1 Since any fractional ideal J is of the form x I with I an integral ideal, decomposing I and (x) into their prime factorizations expresses J as a Z-linear combination of prime ideals, so Frac(R) = G. Corollary 443 allows us to extend the definition of ordp to any fractional R-ideal. Since for a Dedekind domain there is no distinction between invertible fractional ideals and all fractional ideals, the Picard group takes an especially simple form: it is the quotient of the free abelian group Frac(R) of all fractional ideals modulo the subgroup Prin(R) = K × /R× of principal fractional ideals. We therefore have a short exact sequence 0 → Prin(R) → Frac(R) → Pic(R) → 0, and also a slightly longer exact sequence 0 → R× → K × → Frac(R) → Pic(R) → 0. Theorem 444. For a Dedekind domain R, TFAE: (i) Pic(R) = 0. (ii) R is a PID. (iii) R is a UFD. (iv) The set of nonprincipal prime ideals is finite. Proof. Evidently each fractional ideal is principal iff each integral ideal is principal: (i) ≡ (ii). Since R has dimension at most one, (ii) ⇐⇒ (iii) by Proposition 380. Evidently (ii) =⇒ (iv), so the interesting implication is that (iv) implies the other conditions. So assume that the set of (nonzero) nonprincipal prime ideals is nonempty but finite, and enumerate them: p1 , . . . , pn . Let I be an integral ideal, and suppose that I = pa1 · · · pan qb1 · · · qbm . m n1 1 (As usual, we allow zero exponents.) By the Chinese Remainder Theorem we may choose an α ∈ R such that ordpi (α) = ai for all i.60 Now consider the fractional ideal (α−1 )I ; it factors as (α−1 )I = qb1 · · · qbm rc1 · · · rcl , m1 1 l where the ri ’s are some other prime ideals, i.e., disjoint from the pi ’s. But all of the (fractional) ideals in the factorization of (α−1 )I are principal, so (α−1 )I = (β ) for some β ∈ K × and then I = (αβ ) is principal! Remark: There exist one-dimensional Noetherian domains with any finite number √ of nonprincipal prime ideals. For instance, the ring√ [ −3] = Z[t]/(t2 + 3) has Z √ exactly one nonprincipal prime p2 := ⟨1 + −3, 1 − −3⟩. 60Note that we want equality, not just ord (α) ≥ a , so you should definitely think about i Pi how to get this from CRT if you’ve never seen such an argument before. COMMUTATIVE ALGEBRA 249 20.3. Local characterization of Dedekind domains. Theorem 445. Let R be an integral domain. a) If R is Dedekind and S is a multiplicative subset, then S −1 R is Dedekind. b) If R is a Dedekind domain and 0 ̸= p is a prime ideal of R, then Rp is a DVR. Proof. The properties of being Noetherian, dimension at most one and integrally closed are all preserved under localization, so part a) is immediate. Similarly, if 0 ̸= p is a prime ideal, then the localization Rp is a local, one-dimensional integrally closed Noetherian domain, hence by Corollary ?? a DVR, establishing b). Exercise: Let R be a Dedekind domain with fraction field K and p a nonzero prime ideal of R. a) Show that the map ordp : K × → Z defined above is nothing else than the discrete valuation corresponding to the localization Rp . b) Conversely, let v : K × → Z be a discrete valuation. Show that the valuation ring Rv = v −1 (N) is the localization of R at some maximal ideal p. Remark: Call a domain R for which Rm is a DVR for all maximal ideals m a locally Dedekind domain.61 20.4. Factorization into primes implies Dedekind. We are now ready for our last – and most difficult – characterization theorem. Theorem 446. Let R be a domain with the property that every nonzero proper integral ideal is a product of prime ideals. Then R is Dedekind. Proof. Step 1: Let p be an invertible prime of R. We show that p is maximal. Let a ∈ R \ p, and suppose that ⟨a, p⟩ R. Let us then write I1 := ⟨a, p⟩ = p1 · · · pm , I2 := ⟨a2 , p⟩ = q1 · · · qn , where the pi and qj are prime ideals. By assumption, I1 p, and, since p is prime, we have also I2 p. Therefore each pi and qj strictly contains p. In the quotient R = R/p we have (a) = aR = p1 · · · pm and (a2 ) = a2 R = q1 · · · qn . The principal ideals (a) and (a2 ) are invertible, and the pi and qj remain prime in the quotient. Therefore, we have q1 · · · qn = p2 · · · pm 2 . 1 Thus the multisets {{q1 , . . . , qn } and {p1 , p1 , . . . , pm , pm }} coincide, and pulling back to R the same holds without the bars. Thus 2 I1 = ⟨a, p⟩2 = p2 · · · p2 = q1 · · · qn = ⟨a2 , p⟩, 1 m so p ⊂ ⟨a, p⟩2 = a2 R + ap + p2 ⊂ aR + p2 . 61In the literature the term “almost Dedekind domain” is used, but we prefer to reserve this for something else.) That non-Noetherian almost Dedekind domains exist is far from obvious; such rings were first constructed by N. Nakano [Nak53]. 250 PETE L. CLARK So if p ∈ p, p = ax + y with x ∈ R, y ∈ p2 , so ax ∈ p, and since a ∈ R \ p, x ∈ p. Thus p ⊂ ap + p2 ⊂ p, so p = ap + p2 . Multiplication by p−1 gives R = a + p, contrary to hypothesis. So p is maximal. Step 2: Let p be any nonzero prime ideal in R, and 0 ̸= b ∈ p. Then p ⊃ bR and bR = p1 · · · pm , with each pi invertible and prime. Thus by Step 1 the pi ’s are maximal. Since p is prime we have p ⊃ pi for some i and then by maximality p = pi , hence p is invertible. Since by assumption every proper integral ideal is a product of primes, we conclude that every integral ideal is invertible, which, by Theorem 438 implies that R is Dedekind. Remark: Theorem 446 was first proved by K. Matusita [Ma44]. It is often omitted in introductory treatments because of its intricate proof, but it is a very enlightening result because it implies that the ideal theory in any non-Dedekind ring will have to be quite different. For instance, any nonmaximal order in a number field will have “bad ideals” that do not factor into primes. We summarize our work on characterizations of Dedekind domains: Theorem 447. Let R be an integral domain which is not a field. TFAE: (i) R is Noetherian, integrally closed, and of Krull dimension one. (ii) Every fractional (equivalently, every integral) R-ideal is invertible. (iii) R is Noetherian, and the localization at every maximal ideal is a DVR. (iv) Every nonzero proper integral ideal factors into a product of prime ideals. (iv′ ) Every nonzerp proper integral ideal factors uniquely into a product of primes. A ring satisfying these equivalent conditions is called a Dedekind domain. 20.5. Generation of ideals in Dedekind domains. Theorem 448. Let R be a Dedekind domain and I a nonzero ideal of R. Then the quotient ring R/I is a principal Artinian ring. ∏r Proof. Write I = i=1 pai . By the Chinese Remainder Theorem, i R/I ∼ = r ∏ R/pai . i i=1 R/pai i Each factor is also a quotient of the localized ring Rp /pai , which shows that i it is Artinian and principal. Finally, a finite product of Artinian (resp. principal ideal rings) remains Artinian (resp. a principal ideal ring). This has the following striking consequence: Theorem 449. (C.-H. Sah) For an integral domain R, TFAE: (i) R is a Dedekind domain. (ii) For any nonzero ideal I of R and any nonzero element a ∈ I , there exists b ∈ I such that I = ⟨a, b⟩. Proof. The direction (i) =⇒ (ii) follows immediately from Theorem 448. Conversely, assume condition (ii) holds. By Theorem 447 it suffices to show that R is Noetherian and that its localization at each nonzero prime ideal p is a DVR. Certainly condition (ii) implies Noetherianity; moreover it continues to hold for nonzero ideals in any localization. So let I be a nonzero ideal in the Noetherian COMMUTATIVE ALGEBRA 251 local domain (Rp , p). It follows that there exists b ∈ p such that p = I p + bRp . By Nakayama’s Lemma, I = bRp , so Rp is a local PID, hence a DVR. Proposition 450. ([J2, Ex. 10.2.11]) Let R be a Dedekind domain, I a fractional ideal of R and J a nonzero integral ideal of R. Then there exists a ∈ I such that aI −1 + J = R. Proof. Let p1 , . . . , ps be the prime ideals of R dividing J . For each 1 ≤ i ≤ r, choose ai ∈ I p1 · · · pr p−1 \ I p1 · · · pr . Put a = a1 + . . . + ar . We claim that aI −1 + J = R. i It is enough to check this locally. For every prime q ̸= pi , we have JRq = Rq . On the other hand, for all 1 ≤ i ≤ r, aI −1 is not contained in pi , so its pushforward to Rpi is all of Rpi . 20.6. Finitely generated modules over a Dedekind domain. The aim of this section is to prove the following important result. Theorem 451. Let M be a finitely generated module over a Dedekind domain R. Then: a) P = M/M [tors] is a finitely generated projective R-module, say of rank r. b) If r = 0 then of course M = M [tors]. If r ≥ 1 then M ∼ M [tors] ⊕ P ∼ M [tors] ⊕ Rr−1 ⊕ I, = = with I a nonzero ideal of R. c) The class [I ] of I in Pic R is an invariant of M . d) There exists N ∈ Z+ , maximal ideals pi and positive integers ni such that M [tors] ∼ = N ⊕ R/pni . i i=1 Much of the content of the main theorem of this section lies in the following converse of Proposition 16b) for finitely generated modules over a Dedekind domain. Theorem 452. For a finitely generated module M over a Dedekind domain, TFAE: (i) M is projective. (ii) M is flat. (iii) M is torsionfree. Proof. Of course (i) =⇒ (ii) =⇒ (iii) for modules over any domain, and we have seen that (i) ≡ (ii) for finitely generated modules over a Noetherian ring. So it suffices to show (iii) =⇒ (i). Suppose R is a Dedekind domain and M is a finitely generated nonzero torsionfree R-module. By Proposition 16c), we may assume that M ⊂ Rn for some n ≥ 1. We prove the result by induction on n. If n = 1, then M is nothing else than a nonzero ideal of R, hence an invertible module by Theorem XX and thus a rank one projective module by Theorem XX. So we may assume that n > 1 and that every finitely generated torsionfree submodule of Rn−1 is projective. Let Rn−1 ⊂ Rn be the span of the first n − 1 standard basis elements. Let πn : Rn → R be projection onto the nth factor, and consider the restriction of πn to M : n 0 → M ∩ Rn−1 → M → πn (M ) → 0. π 252 PETE L. CLARK Put I = πn (M ). Then I is an ideal of R, hence projective, so the sequence splits: M → (M ∩ Rn−1 ) ⊕ I. Now M ∩ Rn−1 is a torsionfree, finitely generated (since M is finitely generated and R is Noetherian) submodule of Rn−1 , hence is projective by induction. Certainly a direct sum of projective modules is projective, so we’re done. The method of proof immediately yields the following important corollary: Corollary 453. Let P be a finitely generated rank r projective module over a ⊕r Dedekind domain R. Then we have a direct sum decomposition P ∼ = i=1 Ii , where each Ii is a nonzero rank one projective module (equivalently, isomorphic to a nonzero ideal of R). Now let M a finitely generated module over the Dedekind domain R. We have the exact sequence 0 → M [tors] → M → M/M [tors] → 0. Put P := M/M [tors]. Then P is finitely generated (clearly) and torsionfree (by Proposition 16a), hence it is a finite rank projective module (by Theorem 452) and the sequence splits: M ∼ M [tors] ⊕ P. = Lemma 454. I1 , . . . , In be fractional ideals in the Dedekind domain R. Then the ⊕n R-modules i=1 Ii and Rn−1 ⊕ I1 · · · In are isomorphic. Proof. We will prove the result when n = 2. The general case follows by an easy induction argument left to the reader. − Choose 0 ̸= a1 ∈ I1 . Applying Proposition 450 with I = I2 and J = a1 I1 1 ⊂ R, − − − that there exists a2 ∈ I2 such that a1 I1 1 + a2 I2 1 = R. That is there exist bi ∈ Ii 1 such that a1 b1 + a2 b2 = 1. The matrix [ b1 −a2 b2 a1 is invertible with inverse A−1 = [ a1 −b2 a2 b1 . For (x1 , x2 ) ∈ I1 I2 , we have y1 = x1 b1 + x2 ∈ R, y2 = −x1 a2 + x2 a1 ∈ I1 I2 . On the other hand, if y1 ∈ R and y2 = c1 c2 ∈ I1 I2 , then x1 = a1 y1 − b2 c1 c2 ∈ I1 , x2 = a2 y1 + b1 c1 c2 ∈ I2 . Thus [x1 x2 ] → [x1 x2 ]A gives an R-module isomorphism from I1 ⊕ I2 to R ⊕ I1 I2 . Thus we may write M = M [tors] ⊕ M/M [tors] ∼ M [tors] ⊕ = r ⊕ Ii = M [tors] ⊕ Rr−1 ⊕ (I1 · · · Ir ), i=1 which establishes Theorem 451a). As for part b) of the theorem, let T be a finitely generated torsion R-module. Note that the statement of the classification is identical to that of finitely generated torsion modules over a PID. This is no accident, as we can easily reduce to the case of a COMMUTATIVE ALGEBRA 253 PID – and indeed to that of a DVR, which we have already proven (Theorem 414). Namely, let I be the annihilator of T , and (assuming T ̸=∏ as we certainly may) 0, ∏r ⊕r r write I = i=1 pai . Then T is a module over R/I ∼ R/ ⊕ pri ∼ i=1 R/pai . = i i i=1 i = r By Exercise X.X in §3.1, T naturally decomposes as T = i=1 Ti , where Ti is a module over R/pai . This gives the primary decomposition of T . Moreover, each Ti i is a module over the DVR Rp , so Theorem 414 applies. Corollary 455. For any Dedekind domain R, the Picard group Pic R is canonically isomorphic to the reduced K0 -group K0 (R). Proof. Let P be a finitely generated projective R-module of rank r ≥ 1. According to Theorem 451c) the monoid of isomorphism classes of finitely generated projective R-modules is cancellative : this means that the canonical map φ : Pic(R) → K0 (R) φ is injective. It follows easily that the composite map Φ : Pic(R) → K0 (R) → K0 (R) is an injection: indeed, for φ(I ) to be killed in K0 (R) but not K0 (R) it would have to be a fractional ideal which has rank zero as an R-module, and there are no such things. Now an arbitrary nonzero finitely generated projective R-module is isomorphic to Rr−1 ⊕ I , hence becomes equal to the class of the rank one module I in K0 (R), so Φ is surjective. To check that it is a homomorphism of groups we may look on a set of generators – namely, the classes of rank one projective modules. Let us use [P ] for the class of the projective module P in K0 (R) and [[P ]] for its image in K0 (R). Then by Lemma 454 we have Φ([I1 ⊗ I2 ]) = [[I1 ⊗ I2 ]] = [[I1 I2 ]] = [[R ⊕ I1 I2 ]] = [[I1 ⊕ I2 ]] = [[I1 ]] + [[I2 ]]. ¨ 21. Prufer domains A Pr¨ fer domain is an integral domain in which each nonzero finitely generated u ideal is invertible. Since principal ideals are invertible, any B´zout domain is a e Pr¨fer domain. u 21.1. Characterizations of Pr¨ fer Domains. u One might be forgiven for thinking the invertibility of finitely generated ideals is a somewhat abstruse condition on a domain. The following result shows that, on the contrary, this determines a very natural class of domains. Theorem 456. (Characterization of Pr¨fer Domains) For an integral domain R, u TFAE: (i) R is a Pr¨fer domain. u (ii) Every nonzero ideal of R generated by two elements is invertible. (iii) Finitely generated ideals are cancellable: if A, B, C are ideals of R and A is finitely generated, then AB = AC =⇒ B = C . (iv) For every nonzero prime ideal p of R, Rp is a valuation ring. (v) For all ideals A, B, C of R, A(B ∩ C ) = AB ∩ AC . (vi) For all ideals A, B of R, (A + B )(A ∩ B ) = AB . (vii) If A and C are ideals of R with C finitely generated and A ⊂ C , then there exists an ideal B of R such that A = BC . 254 PETE L. CLARK (vii) For all ideals A, B, C of R with C finitely generated, we have (A + B : C ) = (A : C ) + (B : C ). (viii) For all ideals A, B, C of R with C finitely generated, we have (C : A ∩ B ) = (C : A) + (C : B ). (ix) For all ideals A, B, C of R, A ∩ (B + C ) = A ∩ B + A ∩ C . Proof. [LM, Thm. 6.6]. Theorem 457. Let R be a domain. a) Suppose R is a GCD-domain. Then R is a Pr¨fer domain iff it is a B´zout u e domain. b) If R is a Pr¨fer domain and a UFD, then it is a PID. u Proof. a) Since principal ideals are invertible, any B´zout domain is a Pr¨fer doe u main. Conversely, suppose R is a GCD-domain and a Pr¨fer domain. Let x, y ∈ R• u and let d be a GCD of x, y . Certainly we have (d) ⊃ ⟨x, y ⟩. Thus ι : ⟨x, y ⟩ → (d) is a homomorphism of R-modules which we want to show is an isomorphism. By the Local-Global Principle for Module Homomorphisms it is enough to show that for all p ∈ Spec R, ιp is an isomorphism of Rp -modules, i.e., ⟨x, y ⟩Rp = ⟨d⟩Rp . By Proposition 356, d is again the GCD of x and y in the valuation ring Rp (equivalently, the valuation of d is the minimum of the valuations of x and y ) so that the principal ideal ⟨x, y ⟩Rp is generated by ⟨d⟩Rp . b) Suppose R is a Pr¨fer UFD. By part a) it is also a B´zout domain and by XXX u e a B´zout UFD is a PID. e Proposition 458. For a Pr¨fer domain R, TFAE: u (i) R is a B´zout domain. e (ii) Pic(R) = 0. Proof. In the Pr¨fer domain R, an ideal I is invertible iff it is finitely generated. u So (i) and (ii) each assert that every finitely generated ideal is principal. Proposition 459. A Pr¨fer domain is integrally closed. u Proof. In Theorem 438 we showed that a domain in which all fractional R-ideals are invertible is integrally closed. In the proof we only used the invertbility of finitely generated fractional ideals, so the argument works in any Pr¨fer domain. u Exercise: Give a proof of Corollary 459 using the local nature of integral closure. 21.1.1. A Chinese Remainder Theorem for Pr¨fer domains. u Recall that we have a Chinese Remainder Theorem which is valid in any ring: Theorem 126. There is however another, “more elementwise” version of the Chinese Remainder Theorem which holds in a domain R iff R is a Pr¨fer domain. u Let R be a ring, let I1 , . . . , In be a finite sequence of ideals in R and let x1 , . . . , xn be a finite sequence of elements in R. We may ask: when is there an element x ∈ R such that x ≡ xi (mod Ii ) for all i? If we assume that the ideals Ii are pairwise comaximal, then this holds in any ring by the Chinese Remainder Theorem (Theorem 126). But suppose we drop COMMUTATIVE ALGEBRA 255 that condition. Then, if such an x exists, we have x − xi ∈ Ii for all i, hence for all i and j , (25) xi − xj = (x − xj ) − (x − xi ) ∈ Ii + Ij . Thus we get a necessary condition (which is vacuous when the Ii ’s are pairwise comaximal). Let us say that a ring has property ECRT (elementwise Chinese Remainder Theorem) if given any ideals I1 , . . . , In and elements x1 , . . . , xn satisfying (25), there exists x ∈ R such that x ≡ xi (mod Ii ) for all i. Exercise: Show that a PID satisfies property ECRT. Theorem 460. For a domain R, TFAE: (i) R satisfies property ECRT. (ii) R is a Pr¨fer domain. u Proof. . . . 21.2. Modules over a Pr¨ fer domain. u Proposition 461. A domain R is a semihereditary iff it is a Pr¨fer domain. u Exercise: Prove Proposition 461. Lemma 462. Let R be a domain, and let M be a finitely generated torsionfree R-module. Then M is a submodule of a finitely generated free module. Proof. Since M is torsionfree, the natural map M → M ⊗R K is injective, and ι : M ⊗R K ∼ K n for some n ∈ N. Since M is finitely generated, there exists = x ∈ R• such that the image of xM in M ⊗R K is contained in Rn , and thus ι ◦ (x•) : M → Rn . Theorem 463. For an integral domain R, TFAE: (i) Every torsionfree R-module is flat. (ii) Every finitely generated torsionfree R-module is projective. (iii) R is a Pr¨fer domain. u Proof. (i) =⇒ (ii): In particular every finitely generated torsionfree R-module is flat. Since R is a domain, by Corollary 309 every finitely generated flat module is projective. (ii) =⇒ (iii): An ideal of a domain is a torsionfree R-module, so if (ii) holds every finitely generated ideal of R is projective, hence invertible. (iii) =⇒ (i): Let R be a Pr¨ fer domain and M a torsionfree R-module. Then u M = limi Mi is the direct limit of its finitely generated submodules, hence a direct − → limit of finitely generated torsionfree modules Mi . By Lemma 462, each Mi is a finitely generated submodule of a free R-module. By Theorems 461 and 70, each Mi is projective, hence flat. Thus M is a direct limit of flat modules, hence is itself a flat module by Corollary 90. 22. Structure of overrings 256 PETE L. CLARK 22.1. Introducing overrings. Let R be an integral domain with fraction field K . By an overring of R we mean a subring of K containing R, i.e., a ring T with R ⊂ T ⊂ K . (We allow equality.) This is standard terminology among commutative algebraists, but we warn that someone who has not heard it before will probably guess incorrectly at its meaning: one might well think that “T is an overring of R” would simply mean that “R is a subring of T ”. The basic idea of the study of overrings is the following: if R is a very simple ring, then (i) each of its overrings will have very nice properties and (ii) the set of all overrings of R will have simple, transparent structure. Example: Suppose R is a DVR. Then R erly in between. K and there are no overrings prop- Example: Suppose R is a PID. Then the overrings of R correspond to the subsets Σ ⊂ MaxSpec R of nonzero prime ideals: every overring is of the form RΣ , in which we invert one (equivalently, all) generators of all nonzero prime ideals p = (p) ∼ lying outside of Σ. Thus we get a canonical isomorphism MaxSpec RΣ → Σ. Note that for every multiplicative subset S of R, the localization S −1 R is an overring of R. As we have just seen, for PIDs the converse is true. So we may ask: Question 5. For which domains R is every overring a localization? 22.2. Overrings of Dedekind domains. Let R be a Dedekind domain with fraction field K : we assume R ̸= K . Let ΣR denote the set of height one primes of R. For W ⊂ ΣR , we define ∩ RW = Rp p∈W and also RW = ∩ Rp . p∈ΣR \W When W = {p} consists of a single element, we we write Rp for R{p} . Lemma 464. a) For all p ∈ ΣR , there exists fp ∈ Rp \ R. b) The mapping W → RW is injective. Proof. a) Choose x ∈ K \ Rp , and let S be the finite set of maximal ideals q distinct from p such that ordq (x) < ∏ For each q ∈ S , let yq ∈ q \ p. Let 0. N = maxq∈S − ordq (x), and put fp = ( q∈S yq )N x. b) Suppose W1 and W2 are distinct subsets of ΣR . After relabelling if necessary, we may assume that there exists p ∈ W2 \ W1 . By part a), there exists fp ∈ Rp \ R and thus fp ∈ RW1 \ RW2 . Proposition 465. Let R be a Dedekind domain with fraction field K , and let T be an overring of R. Write ι : R → T for the inclusion map. a) For every P ∈ ΣT , T = RP∩R . COMMUTATIVE ALGEBRA b) c) d) e) T is itself a Dedekind domain. ι∗ : ΣT → ΣR is an injection. For all P ∈ ΣT , ι∗ ι∗ P = P . ι∗ identifies ΣT with the subset of p ∈ ΣR such that pT 257 T. Proof. a) Put p = P ∩ R. There exist x, y ∈ R• such that ∈ P . Then 0 ̸= x = y ( x ) ∈ p, so p is a nonzero prime ideal of R. Thus TP contains the DVR Rp and is y properly contained in its fraction field, so TP = Rp . b) By the Krull-Akizuki Theorem, T is a Noetherian domain of Krull dimension one. By part a), the localization of T at every prime is a DVR. So T is integrally closed and is thus a Dedekind domain. c) For distinct P1 , P2 ∈ ΣT , the localizations TP1 , TP2 are distinct DVRs. But by part a), putting pi = ι∗ (Pi ), we have Rpi = TPi for i = 1, 2, so p1 ̸= p2 . d) Suppose p = ι∗ (P ). Then x y (ι∗ ι ∗ P )TP = pTP = P TP . ∗ By part c), ι ∗ ι P is not divisible by any prime other than P , so ι∗ p = P . e) This follows immediately and is stated separately for later use. Lemma 466. Let R be an integrally closed domain with fraction field K , and let T be an overring of R. a) The relative unit group T × /R× is torsionfree. b) Suppose that R is a Dedekind domain, p ∈ ΣR and T = Rp . TFAE: (i) T × /R× ∼ Z. = (ii) T × R× . (iii) [p] ∈ Pic(R)[tors]. (iv) There exists x ∈ R which is contained in p and in no maximal ideal q ̸= p. Proof. a) Since R is integrally closed, all finite order elements of K × (i.e., roots of unity in K ) lie in R and a fortiori in T : R× [tors] = T × [tors]. On the other hand, let x ∈ T × be of infinite order such that xn ∈ R× for some n ∈ Z+ . Again integral closure of R implies x ∈ R, and then xn ∈ R× =⇒ x ∈ R× . b) (i) =⇒ (ii) is clear. (ii) =⇒ (iii): Let x ∈ K × . Then x ∈ T × iff Rx = pa for some a ∈ Z, and x ∈ R× iff a = 0. Therefore (ii) holds iff some power of p is principal, which is to say that the class of p ∈ Pic R is torsion. (iii) =⇒ (i): Let a be the least positive integer such that pa is principal. Thus pa = xR with x uniquely determined modulo R× . It follows that T × is generated by R× and x, so T × /R× is a nontrivial cyclic group. By part a) it is also torsionfree so T × /R× ∼ Z. = (iii) =⇒ (iv): If pa = xR, then x lies in p but in no other maximal ideal q. (iv) =⇒ (iii): If a = vp (x), then a > 0 and xR = pa . Remark: Part (iv) of Lemma 466 was added following an observation of H. Knaf. Theorem 467. (Goldman [Gol64]) For a Dedekind domain R, TFAE: (i) Pic R is a torsion group. (ii) Every overring of R is a localization. Proof. (i) =⇒ (ii): Let p ∈ ΣR . By Lemma 464 Rp is a proper overring of R, so by assumption Rp is a localization of R and thus has a strictly larger unit group. By Lemma 466 this implies that [p] ∈ Pic(R)[tors]. Since Pic(R) is generated by 258 PETE L. CLARK the classes of the nonzero prime ideals, it follows that Pic R is torsion. (ii) =⇒ (i): Let T be an overring of R, and put S = R ∩ T × . We want to show that T = S −1 R. That S −1 R ⊂ T is clear. Conversely, let x ∈ T , and write xR = ab−1 with a, b coprime integral ideals of R: a + b = R. Thus aT + bT = T whereas aT = xbT ⊂ bT , so bT = T and hence also bn T = T for all n ∈ Z+ . Since Pic R is torsion, there exists n ∈ Z+ with bn = bR. It follows that bT = T and thus b ∈ S . Now xb = a ⊂ R, so xb ∈ R. Thus x ∈ S −1 R, and we conclude T ⊂ S −1 R. The last result that we want to prove about overrings of Dedekind domains is a complete classification: let T be an overring of the Dedekind domain R. As we have seen, pulling back via the inclusion ι allows us to view the set of maximal ideals of T as a subset, say W of the set ΣR of maximal ideals of R. Moreover, for each W ⊂ ΣR there is at least one overring T with ΣT = W , namely RW . We claim that in fact these are all the overrings of R, or more succinctly: Theorem 468. For any overrring T of the Dedekind domain R, we have T = Rι∗ (ΣT ) . The proof (that I know) of this theorem requires the study overrings of more general P¨ufer domains, to which we turn in the next section. r 22.3. Overrings of Pr¨ fer Domains. u Theorem 469. Let R be an integral domain with fraction field K , and let T be an overring of R. TFAE: (i) For every prime ideal p of R, either pT = T or T ⊂ Rp . (ii) For every x, y ∈ K × with x ∈ T , ((y ) : (x))T = T . y (iii) T is a flat R-algebra. Proof. . . . Proposition 470. Let R be an integral domain with fraction field K and consider rings R ⊂ T ⊂ T ′ ⊂ K . a) If T ′ is flat over R, then T ′ is flat over T . b) If T ′ is flat over T and T is flat over R, then T ′ is flat over R. c Proof. a) Suppose T ′ is flat over R. Let a, b ∈ T be such that a ∈ T ′ . Write a = s , b c b = d with c, d, s ∈ R. Then d ∈ T ′ , so by Theorem 469, ((d) : (c))T ′ = T ′ . Hence s 1 = t1 u1 + . . . + tk uk for some ti ∈ T ′ and ui ∈ R with ui c ∈ (d) for all i. Then there exists zi ∈ R such that ui c = dzi , so ui a = zi d = zi b ∈ T b for all i. So s (T b : T a)T ′ = T ′ . Applying Theorem 469 again, we conclude that T ′ is flat over T . b) This holds for any tower of ring extensions R1 ⊂ R2 ⊂ R3 , since M ⊗R1 R3 ∼ = (M ⊗R1 R2 ) ⊗R2 R3 . Proposition 471. For an overring T of an integral domain R, TFAE: (i) T is flat over R. (ii) For all p ∈ MaxSpec T , Tp = Rp∩R . ∩ (iii) T = p∈MaxSpec T Rp . Proof. . . . Proposition 472. Let T be an overring of a domain R which is both integral and flat over R. Then R = T . COMMUTATIVE ALGEBRA 259 Proof. Let x, y ∈ R be such that x ∈ T . Then by Theorem 469, ((y ) : (x))T = T . y Let p ∈ MaxSpec R. By Theorem 322, there exists a prime (in fact maximal by Corollary 325, but this is not needed here) ideal P of T lying over p. Since pT ⊂ P , we have pT T . Therefore ((y ) : (x)) is not contained in any maximal ideal of R, so ((y ) : (x)) = R. It follows that x ∈ (y ), i.e., x = ay for some a ∈ R, so that x y ∈ R. Thus R = T . Theorem 473. For an integral domain R, TFAE: (i) R is a Pr¨fer domain. u (ii) Every overring of R is a flat R-module. Proof. (i) =⇒ (ii): An overring is a torsionfree R-module, so this follows immediately from Theorem 463. (ii) =⇒ (i): Let p be a maximal ideal of R. By Proposition 470, every overring of Rp is flat. Let a, b ∈ Rp , and suppose that aRp ̸⊂ bRp . Then (bRp : aRp ) ̸= Rp ; since Rp is local, this implies (bRp : aRp ) ⊂ pRp . Now consider the ring Rp [ a ]. b This is an overring of Rp , hence flat. Since a ∈ Rp [ a ] we have by Theorem 469 b b a a (bRp : aRp )Rp [ ] = Rp [ ]. b b Thus there exist elements x1 , . . . , xn ∈ (bRp : aRp and b1 , . . . , bn ∈ Rp [ a ] such that b x1 b1 + . . . + xn bn = 1. . . . Corollary 474. Every overring of a Pr¨fer domain is a Pr¨fer domain. u u Proof. Let R be a Pr¨fer domain and T be an overring of R. Then every overring u T ′ of T is in particular an overring of R, so T ′ is flat over R. By Proposition ??a), T ′ is also flat over T . Therefore every overring of T is flat over T , so by Theorem 473, T is a Pr¨fer domain. u Finally we give a result which generalizes the (as yet unproven) Theorem 468. Namely, for R a domain, let W be a subset of MaxSpec R and put ∩ RW = Rp . p∈W Theorem 475. Let R be a Pr¨fer domain, T an overring of R, and put u W = {p ∈ MaxSpec(R) | pT T }. Then T = RW . Proof. . . . 22.4. Kaplansky’s Theorem (III). Theorem 476. (Kaplansky [K]) Let R be a Dedekind domain with fraction field K , and let K be an algebraic closure of K . Suppose that for every finite extension L/K , the Picard group of the integral closure RL of R in L is a torsion abelian e group. Then the integral closure S of R in K is a B´zout domain. Proof. Let I = ⟨a1 , . . . , an ⟩ be a finitely generated ideal of S . Then L = K [a1 , . . . , an ] is a finite extension of K . Let RL be the integral closure of R in L, and let IL = ⟨a1 , . . . , an ⟩RL . By hypothesis, there exists k ∈ Z+ and b ∈ RL such that k IL = bRL . Let c be a k th root of b in S and let M = L[c]. Thus in the Dedekind domain RM we have (IL RM )k = (ck ), and from unique factorization of ideals we deduce IL RM = cRM . Thus I = IL RM S = cRM S = cS is principal. 260 PETE L. CLARK Recall the basic fact of algebraic number theory that for any number field K , the Picard group of ZK is finite. This shows that the ring R = Z satisfies the hypotheses of Theorem 476. We deduce that the ring of all algebraic integers Z is a B´zout e domain: Theorem 129. Exercise X.X: State a function field analogue of Theorem 129 and deduce it as a special case of Theorem 476. 22.5. Every commutative group is a Picard group. 23. Krull domains Let R be an integral domain with fraction field K . Let Σ(R) be the set of height one prime ideals of R. We say that R is a Krull domain if it satisfies the following properties: (KD1) For each p ∈ Σ(R), Rp is a DVR. ∩ (KD2) R = p∈Σ(R) Rp . (KD3) For 0 ̸= x ∈ R, {p ∈ Σ(R) | f ∈ p} is finite. In general, let K be a field, and let {Ri } be a family of subrings of K , and put ∩ R = i Ri . We say that the family has finite character (FC) if for any 0 ̸= x ∈ R, × x ∈ Ri for all but finitely many i ∈ I . Exercise X.X: If {Ri }i∈I is a finite character family of subrings of a field K and J ⊂ I , show that the subfamily {Ri }i∈J also has finite character. Theorem 477. For a domain R, TFAE: (i) R is a Krull domain. (ii) There exists a field K containing R, and a finite character family of DVR’s ∩ {Vi } inside K such that R = i Vi . Proposition 478. a) A Krull domain is normal. b) Any normal Noetherian domain is a Krull domain. Proof: a) By (KD1) and (KD2), R is the intersection inside its field of fractions of a family of normal domains, hence it is itself normal. b) . . . Proposition 479. Let R be a Krull domain with fraction field K , and let k be a subfield of K . Then R ∩ k is a Krull domain. Proof: . . . Proposition 480. Let R be a Krull domain with fraction field K , and let L/K be a finite extension. Then the normalization of R in L is a Krull domain. Proof: . . . Proposition 481. Let {Ri }i∈I be a family of Krull subdomains of the field K . If ∩ the family satisfies the finite character property, then R = i Ri is a Krull domain. If R is a Krull domain, with set of height ∩ primes Σ(R), and T ⊂ Σ(R) is an one arbitrary subset, consider the ring RT := p∈T Rp . Evidently RT is an overring of R, i.e., is intermediate between R and its field of fractions. By Exercise X.X, COMMUTATIVE ALGEBRA 261 the family {Rp }p∈T also has finite character, so by Theorem 477 RT is also a Krull domain. An overring of a Krull domain arising from a subfamily T in this way is called a subintersection. In a Krull domain, localizations are subintersections: Proposition 482. Let R be a Krull domain, and S a multiplicatively closed subset. Let T ⊂ ∩ R) be the set of height one primes p which are disjoint from S ; then Σ( S −1 R = p∈T Rp . In particular, a localization of a Krull domain is a Krull domain. Theorem 483. A polynomial ring in any (finite or infinite) number of indeterminates over a Krull domain is a Krull domain. In particular, a polynomial ring in infinitely many indeterminates over a field is a non-Noetherian Krull domain. 24. Connections with model theory References [Ad62] J.F. Adams, Vector fields on spheres. Ann. of Math. (2) 75 (1962), 603-632. [AHRT02] J. Ad´mek, H. Herrlich, J. Rosick´ and W. Tholen, Injective hulls are not natural. a y Algebra Universalis 48 (2002), 379-388. [An00] D.D. Anderson, GCD domains, Gauss’ lemma, and contents of polynomials. NonNoetherian commutative ring theory, 1-31, Math. Appl., 520, Kluwer Acad. Publ., Dordrecht, 2000. E. Artin, Zur Theorie der hyperkomplexen Zahlen. Abh. Hamburg, 5 (1927), 251-260. [Ar27] [AT51] E. Artin and J.T. Tate, A note on finite ring extensions. J. Math. Soc. Japan 3 (1951), 74-77. [At89] M.F. Atiyah, K -theory. Notes by D. W. Anderson. Second edition. Advanced Book Classics. Addison-Wesley Publishing Company, Advanced Book Program, Redwood City, CA, 1989. [AM] M.F. Atiyah and I.G. Macdonald, Introduction to commutative algebra. AddisonWesley Publishing Co., Reading Mass.-London-Don Mills, Ont. 1969. [AuBu59] M. Auslander and D.A. Buchsbaum, Unique factorization in regular local rings. Proc. Nat. Acad. Sci. U.S.A. 45 1959 733-734. [Bae40] R. Baer, Abelian groups that are direct summands of every containing abelian group. Bull. Amer. Math. Soc. 46 (1940), 800-806. H. Bass, Global dimension of rings, Ph.D. Thesis, University of Chicago, 1959. [Bas59] [Bas63] H. Bass, Big projective modules are free. Illinois J. Math. 7 1963 24-31. [Bau62] G. Baumslag, On abelian hopfian groups. I. Math. Z. 78 1962 5354. [Bau63] G. Baumslag, Hopficity and abelian groups. 1963 Topics in Abelian Groups (Proc. Sympos., New Mexico State Univ., 1962) pp. 331-335 Scott, Foresman and Co., Chicago, Ill. [Bor50] S. Borofsky, Factorization of polynomials. Amer. Math. Monthly 57 (1950), 317–320. R. Bott and J. Milnor, On the parallelizability of the spheres. Bull. Amer. Math. Soc. [BM58] 64 (1958), 87-89. [B] N. Bourbaki, Commutative algebra. Chapters 17. Translated from the French. Reprint of the 1989 English translation. Elements of Mathematics (Berlin). Springer-Verlag, Berlin, 1998. A. Bouvier, Le groupe des classes de l’algbre affine d’une forme quadratique. Publ. [Bou78] D´p. Math. (Lyon) 15 (1978), no. 3, 53–62. e D.A. Buchsbaum, Some remarks on factorization in power series rings. J. Math. [Buc61] Mech. 10 1961 749–753. [But64] H.S. Butts, Unique factorization of ideals into nonfactorable ideals. Proc. Amer. Math. Soc. 15 (1964), 21. H.S. Butts and L.I. Wade, Two criteria for Dedekind domains. Amer. Math. Monthly [BW66] 73 (1966), 14–21. 262 [CE] [CaEv59] [Cl66] [Coh68] [Coh73] [CK46] [CK51] [CS46] [Cor64] [Da09] [De71] [DM71] [Ec00] [ES53] [Eis] [FT] [FW] [Fo73] [Fu59] [Fu74] [Ga71] [GJ76] [Gol51] [Gol64] [Gov65] [Gr74] [Hai94] [Has28] [Hel40] [Hei70] [Hi90] PETE L. CLARK H. Cartan and S. Eilenberg, Homological algebra. Princeton University Press, Princeton, N. J., 1956. E.D. Cashwell and C.J. Everett, The ring of number-theoretic functions. Pacific J. Math. 9 (1959) 975–985. L.E. Claborn, Every abelian group is a class group. Pacific J. Math. 18 (1966), 219-222. P.M. Cohn, B´zout rings and their subrings. Proc. Cambridge Philos. Soc. 64 (1968), e 251–264. P.M. Cohn, Unique factorization domains. Amer. Math. Monthly 80 (1973), 1–18. I.S. Cohen and I. Kaplansky, Rings with a finite number of primes. I. Trans. Amer. Math. Soc. 60 (1946), 468-477. I.S. Cohen and I. Kaplansky, Rings for which every module is a direct sum of cyclic modules. Math. Z. 54 (1951), 97-101. I.S. Cohen and A. Seidenberg, Prime ideals and integral dependence. Bull. Amer. Math. Soc. 52 (1946), 252-261. A.L.S. Corner, On a conjecture of Pierce concerning direct decompositions of Abelian groups. 1964 Proc. Colloq. Abelian Groups (Tihany, 1963) pp. 43-48 Akad´miai Kiad´, e o Budapest. C.S. Dalawat, Wilson’s theorem. J. Th´or. Nombres Bordeaux 21 (2009), 517-521. e R. Dedekind. . . F. DeMeyer and E. Ingraham, Separable algebras over commutative rings. Lecture Notes in Mathematics, Vol. 181 Springer-Verlag, Berlin-New York 1971. O. Echi, A topological characterization of the Goldman prime spectrum of a commutative ring. Comm. Algebra 28 (2000), 2329-2337. ¨ B. Eckmann and A. Schopf, Uber injektive Moduln. Arch. Math. (Basel) 4 (1953), 75-78. D. Eisenbud, Commutative algebra. With a view toward algebraic geometry. Graduate Texts in Mathematics, 150. Springer-Verlag, New York, 1995. Field Theory, notes by P.L. Clark, available at http://www.math.uga.edu/∼pete/FieldTheory.pdf C. Faith and E.A. Walker, Direct-sum representations of injective modules. J. Algebra 5 (1967), 203-221. R.M. Fossum, The divisor class group of a Krull domain. Ergebnisse der Mathematik und ihrer Grenzgebiete, Band 74. Springer-Verlag, New York-Heidelberg, 1973. L. Fuchs, The existence of indecomposable abelian groups of arbitrary power. Acta Math. Acad. Sci. Hungar. 10 (1959) 453-457. L. Fuchs, Indecomposable abelian groups of measurable cardinalities. Symposia Mathematica, Vol. XIII (Convegno di Gruppi Abeliani, INDAM, Rome, 1972), pp. 233244. Academic Press, London, 1974. T.E. Gantner, A Regular Space on which every Continuous Real-Valued Function is Constant. Amer. Math. Monthly 78 (1971), 52–53. L. Gillman and M. Jerison, Rings of continuous functions. Reprint of the 1960 edition. Graduate Texts in Mathematics, No. 43. Springer-Verlag, New York-Heidelberg, 1976. O. Goldman, Hilbert rings and the Hilbert Nullstellensatz. Math. Z. 54 (1951). 136-140. O. Goldman, On a special class of Dedekind domains. Topology 3 (1964) suppl. 1, 113-118. ˇ V.E. Govorov, On flat modules. (Russian) Sibirsk. Mat. Z. 6 (1965), 300-304. A. Grams, Atomic rings and the ascending chain condition for principal ideals. Proc. Cambridge Philos. Soc. 75 (1974), 321–329. B. Haible, Gauss’ Lemma without Primes, preprint, 1994. ¨ H. Hasse, Uber eindeutige Zerlegung in Primelemente oder in Primhauptideale in Integrittsbereichen. J. reine Angew. Math. 159 (1928), 3–12. O. Helmer, Divisibility properties of integral functions. Duke Math. J. 6 (1940), 345356. W. Heinzer, Quotient overrings of an integral domain. Mathematika 17 (1970), 139– 148. D. Hilbert, Ueber die Theorie der algebraischen Formen. Mathl Annalen 36 (1890), 473-534. COMMUTATIVE ALGEBRA [Ho69] [Hi75] [H¨01] o [Hun68] [Hus66] [J1] [J2] [Ka52] [Ka58] [K] [Ka08] [Kr31] [Kr37] [Kr51] [Lam99] [Lam05] [Lam06] [Lan02] [LM] [Las05] [Laz64] [Lev43] [Les67] [Li33] [Mal48] [Man58] [Mar71] [M] [Ma44] [Mi65] [Nag57] 263 M. Hochster, Prime ideal structure in commutative rings. Trans. Amer. Math. Soc. 142 (1969), 43–60. J.-J. Hiblot, Des anneaux euclidiens dont le plus petit algorithme n’est pas ` valeurs a finies. C. R. Acad. Sci. Paris S´r. A-B 281 (1975), no. 12, Ai, A411–A414. e O. H¨lder, Die Axiome der Quantitat und die Lehre vom Mass. Ber. Verh. Sachs. Ges. o Wiss. Leipzig, Math.-Phys. Cl. 53 (1901), 1-64. T.W. Hungerford, On the structure of principal ideal rings. Pacific J. Math. 25 (1968), 543–547. D. Husem¨ller, Fibre bundles. McGraw-Hill Book Co., New York-London-Sydney 1966. o N. Jacobson, Basic algebra. I. Second edition. W. H. Freeman and Company, New York, 1985. N. Jacobson, Basic algebra. II. Second edition. W. H. Freeman and Company, New York, 1989. I. Kaplansky, Modules over Dedekind rings and valuation rings. Trans. Amer. Math. Soc. 72 (1952), 327–340. I. Kaplansky, Projective modules. Ann. of Math. 68 (1958), 372-377. I. Kaplansky, Commutative rings. Allyn and Bacon, Inc., Boston, Mass. 1970. M. Karoubi, K-theory. An introduction. Reprint of the 1978 edition. With a new postface by the author and a list of errata. Classics in Mathematics. Springer-Verlag, Berlin, 2008. W. Krull, Allgemeine Bewertungstheorie. J. Reine Angew. Math. 167 (1931), 160-196. W. Krull, Beitr¨ge zur Arithmetik kommutativer Integrit¨tsbereiche, III, zum Dimena a sionsbegriff der Idealtheorie, Mat. Zeit 42 (1937), 745–766. W. Krull, Jacobsonsche Ringe, Hilbertscher Nullstellensatz Dimensionen theorie. Math. Z. 54 (1951), 354-387. T.Y. Lam, Lectures on modules and rings. Graduate Texts in Mathematics, 189. Springer-Verlag, New York, 1999. T.Y. Lam, Introduction to quadratic forms over fields. Graduate Studies in Mathematics, 67. American Mathematical Society, Providence, RI, 2005. T.Y. Lam, Serre’s problem on projective modules. Springer Monographs in Mathematics. Springer-Verlag, Berlin, 2006. S. Lang, Algebra. Revised third edition. Graduate Texts in Mathematics, 211. Springer-Verlag, New York, 2002. M.D. Larsen and P.J. McCarthy, Multiplicative theory of ideals. Pure and Applied Mathematics, Vol. 43. Academic Press, New York-London, 1971. E. Lasker, Zur Theorie der Moduln und Ideale. Math. Ann. 60 (1905), 19-116. D. Lazard, Sur les modules plats. C. R. Acad. Sci. Paris 258 (1964), 6313-6316. F.W. Levi, Contributions to the theory of ordered groups. Proc. Indian Acad. Sci., Sect. A. 17 (1943), 199-201. L. Lesieur, Divers aspects de la th´orie des id´aux d’un anneau commutatif. Enseignee e ment Math. 13 (1967), 75-87. F.A Lindemann, The Unique Factorization of a Positive Integer. Quart. J. Math. 4, 319-320, 1933. A.I. Malcev, On the embedding of group algebras in division algebras (Russian), Dokl. Akad. Nauk. SSSR 60 (1948), 1499–1501. H.B. Mann, On integral bases. Proc. Amer. Math. Soc. 9 (1958), 167–172. C.F. Martin, Unique factorization of arithmetic functions. Aequationes Math. 7 (1971), 211. H. Matsumura, Commutative ring theory. Translated from the Japanese by M. Reid. Second edition. Cambridge Studies in Advanced Mathematics, 8. Cambridge University Press, Cambridge, 1989. ¨ K. Matusita, Uber ein bewertungstheoretisches Axiomensystem f¨r die Dedekindu Noethersche Idealtheorie. Jap. J. Math. 19 (1944), 97–110. J.W. Milnor, Topology from the differentiable viewpoint. Based on notes by David W. Weaver. The University Press of Virginia, Charlottesville, Va. 1965. M. Nagata, A remark on the unique factorization theorem. J. Math. Soc. Japan 9 (1957), 143–145. 264 [Nak53] [Neum49] [Neus07] [No21] [No26] [O74] [Pa59] [Pe04] [Que96] [Qui76] [Ra30] [R] [Ro67] [R¨ 33] u [Ru87] [Sa61] [Sa64] [Sa68] [Se56] [Su76] [Sw62] [Sw69] [T88] [vdW39] [Wa] [Wed07] [W] [Wel80] [Zar47] [Zas69] [Ze34] PETE L. CLARK N. Nakano, Idealtheorie in einem speziellen unendlichen algebraischen Zahlk¨rper. J. o Sci. Hiroshima Univ. Ser. A. 16: 425-439. B.H. Neumann, On ordered division rings. Trans. Amer. Math. Soc. 66 (1949), 202– 252. M.D. Neusel, Invariant theory. Student Mathematical Library, 36. American Mathematical Society, Providence, RI, 2007. E. Noether, Idealtheorie in Ringberereichen. Math. Ann. 83 (1921), 24–66. E. Noether, Der Endlichkeitsatz der Invarianten endlicher linearer Gruppen der Charakteristik p. Nachr. Ges. Wiss. Gttingen: 2835. T. Ogoma, On a problem of Fossum. Proc. Japan Acad. 50 (1974), 266–267. Z. Papp, On algebraically closed modules. Publ. Math. Debrecen 6 (1959), 311-327. H. Perdry, An elementary proof of Krull’s intersection theorem. Amer. Math. Monthly 111 (2004), 356-357. C.S. Queen, Factorial domains. Proc. Amer. Math. Soc. 124 (1996), no. 1, 11–16. D. Quillen, Projective modules over polynomial rings.Invent. Math. 36 (1976), 167171. J.L. Rabinowitsch, Zum Hilbertschen Nullstellensatz. Math. Ann. 102 (1930), 520. M. Reid, Undergraduate commutative algebra. London Mathematical Society Student Texts, 29. Cambridge University Press, Cambridge, 1995. A. Robinson, Non-standard theory of Dedekind rings. Nederl. Akad. Wetensch. Proc. Ser. A 70=Indag. Math. 29 (1967), 444–452. W. R¨ckert, Zum Eliminationsproblem der Potenzreihenideale, Math. Ann. 107 u (1933), 259–281. W. Rudin, Real and complex analysis. Third edition. McGraw-Hill Book Co., New York, 1987. P. Samuel, On unique factorization domains. Illinois J. Math. 5 (1961), 1–17. P. Samuel, Lectures on unique factorization domains. Notes by M. Pavaman Murthy. Tata Institute of Fundamental Research Lectures on Mathematics, No. 30 Tata Institute of Fundamental Research, Bombay 1964. P. Samuel, Unique factorization. Amer. Math. Monthly 75 (1968), 945–952. A. Seidenberg, Some remarks on Hilbert’s Nullstellensatz. Arch. Math. (Basel) 7 (1956), 235-240. A.A. Suslin, Projective modules over polynomial rings are free. Dokl. Akad. Nauk SSSR 229 (1976), 1063-1066. R.G. Swan, Vector bundles and projective modules. Trans. Amer. Math. Soc. 105 (1962), 264-277. R.G. Swan, Invariant rational functions and a problem of Steenrod. Invent. Math. 7 (1969), 148-158. H.F. Trotter, An overlooked example of nonunique factorization. Amer. Math. Monthly 95 (1988), no. 4, 339–342. B.L. van der Waerden, Einftihrung in die algebraische Geometrie, Berlin, 1939. R.B. Warfield, Jr., Rings whose modules have nice decompositions. Math. Z. 125 (1972) 187-192. J.H.M. Wedderburn, On Hypercomplex Numbers. Proc. of the London Math. Soc. 6 (1907), 77118. C.A. Weibel, An introduction to homological algebra. Cambridge Studies in Advanced Mathematics, 38. Cambridge University Press, Cambridge, 1994. R.O. Wells Jr., Differential analysis on complex manifolds. Second edition. Graduate Texts in Mathematics, 65. Springer-Verlag, New York-Berlin, 1980. O. Zariski, A new proof of Hilbert’s Nullstellensatz. Bull. Amer. Math. Soc. 53 (1947), 362-368. H. Zassenhaus, On Hensel factorization. I. J. Number Theory 1 (1969), 291–311. E. Zermelo, Elementare Betrachtungen zur Theorie der Primzahlen. Nachr. Gesellsch. Wissensch. Gttingen 1, 43-46, 1934. ...
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